
Book -^f > y- 

Coipglrtl^? 






COPYRIGHT DEPOSrr. 



ELEMENTARY ELECTRICAL 
CALCULATIONS 



A MANUAL OF SIMPLE ENGINEERING MATHEMATICS, 

COVERING THE WHOLE FIELD OF DIRECT CURRENT 

CALCULATIONS, THE BASIS OF ALTERNATING 

CURRENT MATHEMATICS, NETWORKS 

AND TYPICAL CASES OF CIRCUITS, 

WITH APPENDICES ON 

SPECIAL SUBJECTS 



BY 

T. O'CONOR SLOANE, A.M., E.M., Ph.D. 

AUTHOR OF 

" ARITHMETIC OF ELECTRICITY," " STANDARD ELECTRICAL DICTIONARY," 

"THE ELECTRICIAN'S HANDY BOOK" 




NEW YORK 
D. VAN NOSTRAND COMPANY 

23 Murray and 27 Warren Sts. 
1909 



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Copyright, 1909, by 
D. VAN NOSTRAND CO. 



^"^r 



Stanbope ipctes 

p. U. GILSOH COIalPAllT 
BOSTON. U.S.A. 



LIBRARY of CONGRESS 
Two CoDies Received 

il,4i»^ /^ }^X6 No. 



3 



PREFACE 



This book is designed to give in simple form what may be 
termed a foundation for the study of electrical calculations. 
The operations described require only an elementary knowl- 
edge of mathematics. It is a feature of electrical science that 
although it is built up on a basis of mathematics, a great 
part of the engineering calculations is comprised within the 
limits of arithmetic, while elementary algebra carries it a long 
way further. The algebra required in every day electrical 
work is so simple that it may be learned in a very short time, 
and it is perfectly fair to say that many use it daily without 
realizing that they do so. Some there are who might even 
be repelled from the subject if told that algebra was required 
in its operations, yet such may employ Ohm's law without 
realizing that it is expressed as an algebraic equation. 

Very few calculations in this book call for the use of higher 
algebra than is involved in the treatment of Ohm's law, and 
the arithmetic employed is simpler than that used in com- 
mercial calculations. Where more advanced algebra is 
required, as in the solution of net-works, the matter is placed 
towards the end. The complex variable and the graphic 
solutions of alternating current problems are not employed, as 
they are outside the scope of the book. 

In the Appendix some illustrations of geonietric, of more 
involved algebraic, and of simple calculus solutions and 
demonstrations are given, which may interest those with some 
knowledge of the higher mathematics. 

iii 



iv PREFACE 

The author desires to express his acknowledgment and 
appreciation of the painstaking work of Mr. Newman D. 
Waffle, which has done much for the accuracy of the text and 
treatment of the subject. 

The kindness of readers who will inform the author of any 
errors or misprints to be corrected in future editions will be 
greatly appreciated. 

T. O^CONOR SLOANE. 
New York, March, 1909. 



CONTENTS 



CHAPTER I. 
Introductory. 

PAGE 

Arithmetic. — Logarithms.— The Slide Rule. — Trigonometry. — Indi- 
cation of Division. — Reciprocals. — Symbols. — Subscripts. — Mean- 
ing of n.— Algebraic Symbols. — Algebra. — Positive and Negative. 
— ^Algebraic Addition. — Algebraic Subtraction. — Coefficients. — 
Functions. — Calculus. — Electric Measurements. — Quantity. — Cur- 
rent. — Electro-motive Force. — Resistance. — The Watt. — Equiva- 
lence of Units. — Numerical Values 1-7 



CHAPTER II. 

Exponential Notation. 

Powers. — Exponents. — Roots. — Symbols of Roots. — Fractional Ex- 
ponents. — Decimal Exponents. — ^Negative Exponents. — Multipli- 
cation of Powers. — Division of Powers. — Extraction of Roots by 
Logarithms. — Powers of Ten. — Changing Exponents. — ^Addition 
and Subtraction of Powers. — Factoring Exponents 8-15 



CHAPTER III. 

Mechanics and Physics. 

Units, Simple and Compound. — Significance of Multiplication and 
Division of Different Kinds of Units. — Centimeter-gram-second 
System. — Mass. — Weight and Gravitation. — Space. — Rate Units. 
— ^Velocity. — Acceleration. — Force. — Dyne. — Gravity. — Weight. — 
Acceleration of Gravitation. — Energy. — Available Energy. — 
Entropy. — Potential Energy. — Kinetic Energy. — ^Varieties of 
Energy. — The Erg. — Power or Activity. — British System of 
Units. — ^Value of Kinetic Energy. — Equivalence of Units. — The 
Volt-Coulomb.— The Joule.— The Watt.— Heat Energy.— Rela- 
tions of Different Units. — Relation of Power Units and Energy 
Units. — Efficiency. — Central or Radiant Force. — Force of a Plane 
on a Point near its Surface. — Theory of Dimensions. — Dimensions 
of Mechanical Units. — Problems 16-40 



vi CONTENTS 

CHAPTER IV. 
Ohm's Law. 

PAGB 

Three Factors of an Active Circuit. — Ohm's Law. — Several Appli- 
ances in One Circuit. — Apphcation of Ohm's Law to Portions of a 
Circuit. — Simple Method of Expressing Ohm's Law. — Propor- 
tional Form of Ohm's Law. — Fall of Potential. — RI Drop. — 
Counter Electro-motive Force. — ^Problems 41-52 



CHAPTER V. 

Resistance. 

Linear Conductors. — Resistance and Weight of Linear Conductors. — 
Parallel Conductors. — Distribution of Current in Parallel Conduc- 
tors. — Resistances of Conductors in Parallel. — Combined Resistance 
of Two Conductors. — Combined Resistance of any Number of 
Conductors. — Specific Resistance. — Circular Mils. — Efifect of 
Temperature of Conductors on their Resistance. — Problems 53-76 



CHAPTER VI. 

Kirchhoff's Laws. 

Kirchhoff's First and Second Laws. — Problems 7 7-81 

CHAPTER VII. 

Arrangement of Batteries. 

Electro-motive Force of a Battery. — Resistance of a Battery. — Poten- 
tial Drop of a Battery. — Greatest Current from a Battery. — Rules 
for Calculating a Battery. — Energy Expended in a Battery. — Rule 
for Calculating a Battery of Given Efficiency. — Discussion. — 
Problems 82-93 



CHAPTER VIII. 

Electric Energy and Power. 

Potential. — Proof of Numerical Value of Potential. — Potential Drop, 
otherwise called RI Drop. — Electric Energy. — Practical Unit of 
Electric Energy. — Electric Power or Activity. — Practical Unit of 
Electric Power. — Relations of Power to Current, Resistance, and 
e.m.f . — Equivalents of the Watt. — Problems 94-104 



CONTENTS vn 

CHAPTER IX. 

Bases and Relations of Electric Units. 

PAGB 

Effects of an Electric Current.— Two Systems of C.G.S. Electric Units. 
— The Basis of Practical Units. — The Basis for Measurement and 
Definition of a Current. — The Absolute C.G.S. Electro-magnetic 
Unit of Current. — The Tangent Compass. — Action of Earth's 
Field. — ^Angle of Divergence. — Combined Action of Coil and 
Earth's Field on a Permanent Magnet. — Tangent Galvanometer 
Formula. — Determining C.G.S. units of e.m.f. and Resistance. — • 
The Absolute C.G.S. Electrostatic Unit of Quantity. — Determina- 
tion of the E.S. Unit of Potential. — The Attracted Disk Electrom- 
eter. — Other Units of the Electrostatic System. — The E.M. and 
E.S. System of Units. — Equivalents of the Two Systems of Units. — 
Derivation of Practical Units from Absolute Units. — Reduction 
Factor for Units. — Dimensions of E.M. Quantities. — Dimensions of 
Magnetic Quantity. — Dimensions of Current in E.M. System. — 
Dimensions of Electric Quantity in E.M. System. — Dimensions of 
Potential in E.M. System. — Dimensions of Resistance in E.M. 
System. — Dimensions of Capacity in E.M. System. — Dimensions 
of Electric Quantity in E.S. System. — Dimensions of Surface Den- 
sity in E.S. System. — Dimensions of Potential in E.S. System.— 
Dimensions of Capacity in E.S. System. — Dimensions of Current 
in E.S. System. — Dimensions of Resistance in E.S. System. — 
Dimensions of Magnetic Quantity. — Dimensions of Surface Den- 
sity of Magnetism. — Dimensions of Magnetic Intensity. — Dimen- 
sions of Magnetic Potential. — Dimensions of Magnetic Power. — 
Dimensions of Electric Intensity in E.S. System. — Problems. . . .105-125 

CHAPTER X. 

Thermo-Electricity. 

Emissivity. — ^Heating of a Conductor by a Current. — Cross Section of 
a Conductor to be Heated to a Given Degree by a Given Current. 
— Thermo-electric Couple. — Electro-motive Force of a Thermo- 
electric Couple. — Neutral Temperature. — Temperature and Electro- 
motive Force of Thermo-electric Couple. — Thermo-electric Tables. 
— Peltier Effect. — Absolute Temperature. — Thomson Effect. — 
Problems 126-141 



CHAPTER XL 

Electro-Chemistry. 

Chemical Composition.--— Chemical Saturation. — Chemical Equiva- 
lents and Atomic Weights. — Electric Decomposition or Electroly- 
sis. — Relation of Chemical Equivalents to Electrolysis. — Electro- 
chemical Equivalents. — Thermo-electric Chemistry. — Calculation 
of Electro-motive Force of a Voltaic Couple. — Problems. 142-154 



VUl CONTENTS 

CHAPTER XII. 

Fields of Force. 

PAGE 

Fields of Force.— The Unit Field.— Intensity of Field.— Polarity. — 
Quantity of Field. — Lines of Force per Square Centimeter and per 
Square Inch. — Kapp's Unit. — Fields of Uniform and of Varying 
Strength. — Radiant Fields of Force. — Reciprocal Action in Fields 
of Force. — Induction of e.m.f. and Current. — Problems 155-165 



CHAPTER XIII. 

Magnetism. 

Moments. — Lever Arm of a Force. — The Couple. — Unit of Moment. 
— The Magnetic Filament. — Lines of Force in a Filament. — 
Magnetic Quantity and Strength of Pole. — Measure of Magnetic 
Quantity. — Lines of Force Produced by Unit Quantity of Magne- 
tism. — Moment of a Magnet. — Intensity of Magnetization. — Two 
Definitions of Intensity of Magnetization. — ^Turning Moment of a 
Magnet. — Magnetic Traction. — Problems 166-177 



CHAPTER XIV. 

Electro-Magnetic Induction. 

Induction of Magnetism. — Relation of Induced Magnetization to 
Field. — Susceptibility. — Table of Susceptibility. — Magnetic Induc- 
tion. — Permeability. — Permeance. — Reluctance. — Permeance and 
Reluctance. — Formulas for Inch Measurements. — The Magnetic 
Circuit. — Ampere Turns. — Intensity or Strength of Field Referred 
to C.G.S. Unit Turns. — Strength of Field Referred to Ampere 
Turns. — Total Field Referred to Ampere Turns. — Reluctance of 
Air. — Magneto-motive Force. — Intensity of Field at Center and 
Ends of Coil Interior. — Magnetic Circuit Calculations. — Reluc- 
tance of Circuit of Iron. — ^Ampere Turns for a Given Field. — 
Magnetic Traction. — Determination of Permeability from Traction. 
— Problems 178-197 



CHAPTER XV. 

Capacity and Inductance. 

Capacity. — Measure of Capacity. — Capacity of Parallel Plates. — 
Equations for Capacity Calculations. — Energy of a Charge. — 
Specific Inductive Capacity. — Measure of Specific Inductive Capa- 



CONTENTS IX 

PAGB 

city. — Relation of Absolute Potential, Capacity and Quantity. — 
Value of Absolute Electric Potential. — Potential Difference and 
Transfer of Quantity. — Heat Analogy of Potential and Capacity. — 
Inductance. — The Henry and Rate of Current Change. — Induc- 
tance Formulas. — Variation of Rate of Current Change. — Energy of 
the Electro-magnetic Field of Force. — Problems 198-210 



CHAPTER XVI. 

Hysteresis and Foucault Currents. 

Hysteresis Loss. — Steinmetz's Hysteresis Formula and Table. — 
Steinmetz's Formula Based on Weight of Iron. — Foucault or Eddy 
Currents. — Formulas for Laminated Cores. — Formulas for Wire 
Cores. — Copper Loss in Transformers. — Efficiency of Transformers. 
— Ratio of Transformation in Transformers. — Problems 211-219 



CHAPTER XVII. 

Alternating Current. 

Induction of Alternating e.m.f. — Alternating Current. — The Sine 
Curve. — Sine Functions. — Cycle. Frequency. — Value of Instan- 
taneous e.m.f. and Current. — Average Value of Sine Functions. — 
Effective Values. — Form Factor. — Reactance of Inductance. — Rate 
of Change. — Deduction of Ohmic Value of Inductance Reactance. 
— Impedance of Inductance and Resistance. — Lag and Lead. — Lag 
of Current. — Deduction of Ohmic Value of Capacity Reactance. — 
Combined Impedance of Inductance and Capacity. — Lead of Cur- 
rent. — Lag or Lead due to both Reactances. — Impedance of Induc- 
tance, Capacity, and Resistance Combined. — Angle of Lead or Lag. 
— Povi^er and Power Factor. — Power Factor for both Reactances 
Combined. — ^Angle of Lag and Rate of Change. — Problems . . . 220-246 



CHAPTER XVIII. 

Networks. 

Cycles and Meshes. — Direction of Real Current. — Outer and Inner 
Meshes. — Indication of Cycles. — Cycle Equations. — Cycle Calcu- 
lations. — Meaning of Cycle Letter. — Currents in a Network. — • 
Maxwell's Rule for Writing the Equations of the Cycles. — Resist- 
ance of a Network. — Fleming's Method of Calculating for Resist- 
ance of a Network. — Problems 247-269 



1 

j 

i 
\ 

X CONTENTS ! 

\ 

CHAPTER XIX. 

J 

Demonstrations by Calculus. 

PAGE J 

Force Exerted by Infinite Plane on a Point at Finite Distance. — j 

Absolute Potential. — ^Average Value of Sine Functions. — Effective 
Value of Sine Functions. — Rate of Change. — Resistances of a ' 

Battery for Maximum Current 270-276 j 



APPENDIX A. 
Geometrical Solution of Parallel Circuits 277-278 

APPENDIX B. 
Algebraic Solution of Circuits 279-286 



APPENDIX C. 
VVheatstone Bridge Law 287-288 ; 



Tables of Equivalents 289-291 

Index 293-304 



ELEMENTARY ELECTRICAL 
CALCULATIONS 

HOW MADE AND APPLIED 



CHAPTER I. 

INTRODUCTORY. 

Arithmetic. — Logarithms. — The Slide Rule. — Trigonometry. — Indi- 
cation of Division. — Reciprocals. — Symbols. — Subscripts. — Mean- 
ing of 11. — Algebraic Symbols. — Algebra. — Positive and Negative. 
— Algebraic Addition. — Algebraic Subtraction. — CoefScients. — 
Functions. — Calculus. — Electric Measurements. — Quantity. — Cur- 
rent. — Electro-motive Force. — Resistance. — The Watt. — Equivalence 
of Units. — Numerical Values. 

Arithmetic. In practical electrical calculations much can 
"be done by arithmetic. The operations may be abridged by 
various special methods applying the properties of numbers. 
The following are examples. 

To multiply by 5, add a cipher and divide by 2. To multiply 
by 25, add two ciphers and divide by 4. To square any even 
number between 12 and 24, square one-half the number and 
multiply by 4. To square any number which is divisible by 3, 
not exceeding 36, square one -third of the number and multiply 
by 9. To multiply numbers which can be factored into small 
factors, the factors can be multiplied together. Thus 63 X 49 
= 7X9X7X7 = 3087. 

There are other methods similar to these which can be 
utilized for mental calculations. 

Special attention should be paid to exponential notation, 
as it plays an important part in dimensional formulas and 
deductions and in notation by powers of 10. 

Cancellation can sometimes be used to great advantage; 

I 



2 ELEMENTARY ELECTRICAL CALCULATIONS 

often it is practically unavailable. Inspection of the problem 
will decide whether it is applicable or not. 

Where decimals enter into a problem it is a question to be 
determined in each case how far they should be carried out. 
In the present book they are carried out far enough for most 
purposes of electrical work, which in this respect is not as a 
rule very exacting. 

Logarithms, Familiarity with logarithms should be 
acquired. In many cases they faciHtate operations; in the 
extraction of roots they are often indispensable. 

The Slide Rule, The slide rule is a favorite instrument 
of the electrical and mechanical engineer. It gives practi- 
cal results with great rapidity, but its limitations should be 
kept in mind. If of pocket size, it gives only approximate 
results, not as close as those given by logarithms, whose prin- 
ciples are carried out by it. 

Trigonometry. Trigonometric functions are used to a 
limited extent in alternating-current work and in deductions of 
laws. Tables of natural sines and tangents are accurate enough 
for most purposes. Some familiarity with analytical trigonome- 
try is desirable, and at least a knowledge of the ordinary 
trigonometric functions should be possessed by the student. 

Indication of Division. There are various ways of indi- 
cating the division of one quantity by another. Suppose the 
division of 4 by 3 is to be indicated, it may be done as follows: 

4 -7- 3, Z, and 4 X 3~^. Each of these expressions may be 

3 
correctly read "four divided by three." This is the best way 
to express a fraction, as it keeps before the mind something 
often inadequately realized, namely, that a fraction is an indica- 
tion of division. The fraction % may be correctly called four- 
thirds; it should never be called " four over three." 
When the constituents of a fraction are symbols the only 



INTRODUCTORY 3 

proper designation is the one employing the words "divided 
by. " — is to be expressed as " a divided by 6," not " a over b." 



Reciprocals. The reciprocal of an integral number is a 
fraction having i for its numerator and the number for its 
denominator. The reciprocal of a fraction is the fraction 
inverted. The reciprocal of 3 is J; the reciprocal of f is |. If 
the quantity is a mixed number reduce to a fraction and invert it. 

Example. What is the reciprocal of (i)— , of (2) 4f, and 
oi(3){x+y)? * 

Solution. (I) K (2) A (4f = J^^), and (3) — ^ • 
a (x+ y) 

Symbols. To avoid the writing out of names, symbols 
are employed, such as e.m.f. for electro-motive force. To a 
considerable extent the same symbols are used by all electri- 
cians. In formulas single -letter symbols are used, such as E 
for electro-motive force. 

Subscripts. When the same letter has to be used for 
different places the places may be designated by letters or 
numbers. Then the letter of the symbol is used with the 
distinguishing character placed at its foot as a subscript. 

Suppose a current of electricity goes through a number of 
conductors designated as i, 2, 3, and so on, and that resistance 
is denoted by the capital letter R. Then the resistance of 
each conductor would be denoted respectively by R^, R^, R^, 
and so on. 

Meaning of n. Sometimes a series of quantities, the num- 
ber of which quantities is undetermined, has to be indicated. 
This is done by the use of the letter n, thus : Suppose resist- 
ances were to be indicated as above ; it would be done thus : 
2?i, R2, i?3 . . . i?„. li n were once determined or settled 
on, a number could be substituted for it. Thus R^, R^, 
-R3 ... i?e means a series of such "i?'s," up to Rq. The 
first or "n series'* means the same up to Rn- 



4 ELEMENTARY ELECTRICAL CALCULATIONS 

Algebraic Symbols. Algebraic symbols are used in the de- 
duction and expression of relations between quantities whose 
numerical values are not expressed or are unknown. Ohm's 
law in one form reads: Current is equal to electro-motive force 
divided by resistance. The algebraic expression of the law is 

-I 

If for any two of these symbols numerical values are substi- 
tuted, the simple arithmetical operation indicated will give the 
value of the third. 

Algebra. Algebra is the shorthand of arithmetic, and 
calculus is the shorthand of algebra. In both branches of 
mathematics the power and scope of the lower ones are also 
extended. While a great part of the necessary operations of 
electrical calculation can be done by arithmetic, some use of 
algebra is to be recommended. Sufficient knowledge of alge- 
bra for most purposes can be acquired in a very few weeks. 

Positive and Negative. Quantities sometimes have posi- 
tive values. Such are written without any sign. In formulas 
involving addition the sign is expressed. If a quantity has a 
negative value the sign must be prefixed, as — 4, —E. Sup- 
pose that two e.m.f.'s were opposed to each other so that one 
tended to produce a current in one direction and the other 
e.m.f. tended to produce a current in the other direction. 
Then we could express this condition of things by giving a 
positive value to one e.m.f. and a negative sign to the other, as 
£and -E. 

Algebraic Addition. Algebraic addition is the adding 
together of quantities, taking their signs into account. To do 
it, when quantities have different signs, add together all quan- 
tities of the same sign, thus getting two sums. Subtract the 
numerically smaller one from the larger one and give the 
remainder the sign of the larger of the two original sums. If 



INTRODUCTORY 5 

all the numbers to be added are of the same sign, add all 
together and prefix the sign of the original number to the sum 
if negative. 

Example. Add together 5, 7, — 4, —2 and 2. 

Solution. Proceeding as above we have 

5 and — 4 and subtracting 14 
7—2 6 

— - 6 8 

14 
to which a positive value is assigned, because the sign of the 
larger of the two sums, 14, is positive. 

Algebraic Subtraction. Algebraic subtraction is done by- 
changing the sign of the number to be subtracted and then 
adding the two algebraically. 

Example. Subtract —9 from — 3. 

Solution. — 9 is the subtrahend, or number to be sub- 
tracted. Changing its sign we have to add 9 and — 3 alge- 
braically to carry out the operation described. The algebraic 
sum is 6, whose sign is positive because the sign of the larger 
of the two quantities is positive. 

In algebraic subtraction it is immaterial whether the num- 
ber to be subtracted is larger or smaller than the number from 
which it is to be subtracted. 

Coefficients. A coefficient is the multiplier of a symbol. 
Suppose the symbol r indicated 4 ohms; then in the expression 
7 r, 7 would be the coefficient of r, and the expression would 
indicate 7X4 ohms = 28 ohms. 

Functions. If a quantity multipHed or divided by a num- 
ber gives a product or quotient equal to another quantity, the 
latter quantity is a function of the first. This is expressed 
algebraically in this way: y =^ f {x\ a = f (6), reading " ); is a 
function of x" and "^ is a function of S.'* Suppose that 
a == ^h, then a would be a function of h. The same would be 



6 ELEMENTARY ELECTRICAL CALCULATIONS 

the case if fl = 6/3. li y = 2 oc^, then y would be a function 
of x^, ory =f (x^). 

Calculus. Many electrical deductions are easily reached by 
calculus. Some of these are given in a separate chapter. 

Electric Measurements. — Quantity. Electricity may be 
measured by any one of its various effects. A quantity of 
electricity which will precipitate 0.001,118 gramme of silver 
is the practical or working measure of electric quantity and 
is called the coulomb. 

Current. Under proper conditions what is called a current 
of electricity passes through a conductor. A current of such 
intensity that one coulomb per second passes is called a cur- 
rent of one ampere strength or intensity. 

A no-volt 16 candle power lamp uses about one-half an 
ampere when burning. 

E.M.F. The conditions for producing a current include 
a conductor, such as a wire, and the maintenance of a poten- 
tial difference between the ends of the conductor. The differ- 
ence of potential is called potential difference, potential drop, 
and electro-motive force. The latter is abbreviated into e.m.f. 
Sometimes the capitals are employed, thus, E.M.F. The 
measure of e.m.f. is the volt for all ordinary purposes of engi- 
neering. 

Resistance. A volt will produce a greater or less current 
through a conductor according to the size and material of the 
conductor. A volt will produce a current of almost one 
ampere through a copper wire No. 10 gauge and 1,000 feet 
long. 

A conductor through which an e.m.f. of one volt will produce 
a current of one ampere is said to have a resistance of one ohm. 

The Watt. If the number of volts acting on a conductor 
are multipHed by the amperes of current produced, the result 
expresses the number of a compound measure present, which 



INTRODUCTORY *j 

measure is called the volt-ampere or watt. The watt is a unit 
of rate of electric energy. 

A watt will raise a pound nearly | foot per second. A watt 
acting for a second is a volt-coulomb, which is the unit of elec- 
tric energy and is equal to a joule. The two names are nearly 
synonymous, and in many cases can be used interchangeably. 

Equivalence of Units. All units of force and energy in 
whatever systems can be reduced one to the other by the use 
of multipliers or divisors, which latter are called equivalents. 
Thus from the table of equivalents we can find the value of the 
watt in heat or in mechanical units and the value of the dyne 
in other units of force. But for the electrical units of current 
and of electro-motive force no equivalents exist. The same, 
with some slight reservation, may be said of other electric units. 

Numerical Values. There are a number of numerical 
values which it is well to memorize. The following are examples. 
Each branch has its own set of constants and values, and it is 
a question of individual memory how many can be learned 
and retained. 
30 feet a second, a single rail length, is about 20 miles an hour. 

60 feet a second, a double or trolley rail length, is about 40 

miles an hour. An ordinary railroad passenger car may be 

taken as 60 feet long for this computation. 
1.47 feet in a second is i mile an hour. 
The acceleration of gravitation is about 32.2 feet, 981 cm. 

\/2 = 1.414; -4=^ o.^o-]. 

V2 

Copper wire i foot long, i mil. diameter, has 10.79 ohn^s 

resistance. 
Copper wire No. 10 A.W.G., 1,000 feet long, has about i ohm 

resistance. 
746 watts = I horse power. 
;r=:3.i4i59; for approximate calculations, 3I, ^^. 

2 ;:== 6.2832; 4;r= 12.5664; - =0.7854; i radian = 57.296°. 

4 



CHAPTER II. 

EXPONENTIAL NOTATION. 

Powers. — Exponents. — ■ Roots. — Symbols of Roots. — Fractional Ex- 
ponents. — Decimal Exponents. — Negative Exponents. — Multipli- 
cation of Powers. — Division of Powers. — Extraction of Roots by 
Logarithms. — Powers of Ten. — Changing Exponents. — Addition 
and Subtraction of Powers. — Factoring Exponents. 

Powers. A power of a number is a number produced by 
multiplying the original number by itself any number of times. 

Exponents. An exponent is the index of a power. It 
indicates multiplication by itself of the quantity to which it is 
affixed, such multiplication to be repeated as many times as 
there are units in the exponent less one. Thus 2^ indicates 
that 2 shall be multiplied by 2 four times. Writing it out it 
becomes 2X2X2X2X2 = 32. It is seen that an expo- 
nent indicates the number of times a number is used as a factor 
in producing another number. In 32 the number 2 appears 
5 times as a factor, hence its exponent is 5. 

Roots. A root of a number is a number which multiplied 
by itself a certain number of times, as above, will produce 
the original quantity. 2 is a root of 4, because if multiplied 
by itself it will produce 4, thus, 2X2=4. A root of a higher 
order than the third root (cube root) is designated numerically ; 
an index number is assigned all roots, which number indicates 
how often the root enters as a factor into the multiplication 
producing the original quantity. Thus 2 is the third root of 
8, because if taken as a factor three times it gives 8, thus, 
2X2X2=8. 

The second root of a quantity is usually called the square root, 
and the third root is usually called the cube root. Other roots 
are specified by number, as the fourth root, fifth root, and so on. 

8 



EXPONENTIAL NOTATION $ 

Symbols of Roots. A root is indicated by the radical 

sign (\/ ) with the index number above the opening when higher 
than the second place. This is the usual method in arithmetic. 
Thus \/4 indicates the square root or the second root of 4, 
namely 2; 'sJSi indicates the fourth root of 81, namely 3, for 

3 X3 X3 X3 =81. But in electrical work the system of 
indicating roots by fractional exponents is generally employed. 
A root is indicated by a fractional exponent whose numerator 
is unity and whose denominator is the index of the root. The 
two roots given above would be indicated thus: 4* = 2 and 
81* = 3. As a matter of nomenclature they may be called 
four to the one-half power and eighty-one to the one-fourth 
power. 

Fractional Exponents. Exponents may be fractional, 
with any number in the numerator. The signification of a frac- 
tional exponent with a numerator of i has been explained. 
A fractional exponent in general indicates the root of a power 
of the number to which it is affixed. The power is the one 
indicated by the numerator of the fraction, the root is the one 
indicated by the denominator of the fraction. 

Example. Calculate the value of 4^. 

Solution. It is the fourth root of the second power of 4, 
or the second power of the fourth root of 4. It is immaterial 
which order is followed. Sometimes one is easier than the 
other. Performing the operations, we have 4^ = 16. This is 
the second power of 4, because 4 X 4 = 16. The fourth root 
of 16 is 2, because 2X2X2X2= 16. Therefore 4? = 2. 

4 may be written in other ways. Its exponent I can be 
reduced to its lowest terms and can be written J or 0.5. 
Substituting these expressions for I gives 44 = 4^ = 4"' =2, 
as before. 

Decimal Exponents. An exponent consisting of a decimal 
fraction can be expressed as a vulgar fraction. Thus 10°'^^^ = 



10 ELEMENTARY ELECTRICAL CALCULATIONS 



loi^^Tfff. From a table of logarithms the above is found to be 
5.012. • 

Negative Exponents. A negative exponent indicates the 
reciprocal of the indicated power of the number. 

Example. Express the value of 5-^ = -3 * 

Solution. 5^ = 125, The reciprocal, -^' is — which is 

the value of 5-^. 

Multiplication of Powers. The product of identical 
numbers with exponents affixed is equal to the number raised 
to the power indicated by the algebraic sum of the exponents. 

Example. Multiply 6^ by 6^. 

Solution. The sum of the exponents is 5. Then 6^ X 6^ = 
6^ = 7776. 

Example. Multiply 6^ by 6~^. 

Solution. The sum of the exponents is i, giving as result 
6^ = 6. 

Example. Multiply 6"^ by 6^. 

Solution. The sum of the exponents is — i, and 6~* =-' 

This can be done in another way. 6-^ = — r • 6^ = 36. 

— c X 3^ = .: = 7 ) as before. 
216 216 6 

Example. Appty the same process to multiplying 6^ by 6"^. 

Solution. 6^ = 216. 6-^ =-—-•—— = 6, as in the exam- 

36 36 

pie above. 

Example. Multiply 32 by 3^. 

Solution. The sum of the exponents is— =2. The product 

2 

is then 3^ = 9. 

Division of Powers. The quotient of identical numbers 
with exponents affixed is equal to the same number with a new 



EXPONENTIAL NOTATION II 

exponent, which is the algebraic difference of the two exponents, 
that of the dividend being taken first as minuend. 

Example. Divide 4^ by 4^ 

Solution. Subtracting the exponents for the new expo- 
nent and aflSxing it to the original number gives 

4 ~ 4 — 4 • 

Doing the same by arithmetical process, we have 4* = 256 
and 4^ = 16. 256 -T- 16 = 16 = 4^, as before. 

Example. Divide 4^ by 4^. 

Solution. Proceeding as above we find for the new expo- 
nent 3 — 4 = — I. The quotient then is the original number, 4, 
with an exponent — i, or 4-^, = J. 

Applying ordinary arithmetic, we have 4^ = 64, 4* = 256, 

and 64 -^ 2tc6 = — - = - , as before. 
^ 256 4 
Example. Divide 5^ by 5*. 

Solution. The quotient is 5. No exponent is expressed 
because it is unity and positive, and such exponent is not 
written out but is understood. 

Extraction of Roots by Logarithms. To extract a root 
use logarithms, except for the square root, whose extraction is 
so simple that it is applicable for computation. In using 
logarithms for this as for other purposes pay strict attention to 
the characteristic. Do not guess at the decimal place. If the 
logarithm to be divided by the index of the root has a negative 
characteristic proceed as follows. Multiply 10 by the divisor, in 
the present case by the index of the root, and place it after the 
mantissa of the logarithm with a negative sign between. Sub- 
tract numerically the characteristic of the logarithm from the 
same product and place it in front of the logarithm as a new 
characteristic in place of the original one, giving it a positive 
sign. 

Example. Divide the logarithm "2.25042 by 7. 



12 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. lo X 7 =70, which is to follow the mantissa 
with a minus sign between. 70—2 =68, which is the new 
characteristic . of the logarithm, and we have the division 
(68.25042 — 70) -T- 7 = 9.75006 — 10 = 1.75006. 

Powers of Ten. Notation by powers of ten is simply a 
special case of exponential notation, subject to the rules as 
given. The exponent of a power of ten indicates the number 
of ciphers there would be in its development. Thus 10^ indi- 
cates 1000, 10^ indicates 1,000,000, and so on. 

Example. Write 34,000,000,000 in powers of ten. 

Solution. It is 34 multiplied by 10 raised to the power 
indicated by the number of ciphers. There are 9 ciphers; 
therefore it is written 34 X 10^. 

Example. Multiply 34 X 10® by 27 X 10^. 

Solution. 34 X 27 X 10^^ = 918 X lo^^ 

For all operations of multipHcation and division in powers of 
ten, follow the rules of ordinary arithmetic as far as the num- 
bers other than the powers of ten are concerned. For the 
powers of ten follow the rules given for exponential notation. 

Example. Divide 34 X 10^ by 27 X lo^ 

Solution. Divide 34 by 27 and multiply the quotient by 10 
raised to the power indicated by the exponent obtained by sub- 
tracting 6 from 9, w^hich is 3. This gives — X lo^ 

27 

Example. Divide 29 X 10^ by 21 X lo^ 

Solution. 29 -T- 21 = 1.38. The new exponent is 7 — 9 
= — 2, giving as the quotient 1.38 X io~^, which, if we depart 
from powers of ten, is 0.0138. 

Changing Exponents. The exponent of the ten factor 
may be changed by changing the location of the decimal point 
of the number. If the exponent is diminished, the decimal 
point of the number must be moved to the right as many points 
as the units removed from the exponent. If the exponent is 



EXPONENTIAL NOTATION 1 3 

increased, the decimal point of the number is moved to the left a 
corresponding number of points. This process enables us to 
remove the decimal point from the number if it is introduced 
by any operation. 

Example. Remove the decimal point from 1.38 X lo^ 

Solution. If the exponent of 10^ is reduced by 2, the decimal 
point of the number must be moved two points to the right. 
This will remove the decimal point. Thus 1.38 X 10^ = 138 
X lo^. 

Example. Reduce 1.996 X lo"^ to an expression without a 
decimal. 

Solution. We see that the exponent of 10 must be dimin- 
ished by 3. This is done by algebraic subtraction. — 7 — 3 
= —10, and 1.996 X 10-^ = 1,996 X lo"^''. 

Addition and Subtraction of Powers. Exponential 
notation applies principally to the operations of multiplication 
and division. Numbers of different exponents cannot be added 
or subtracted. If such operation is to be done, the numbers 
must be reduced to factors of the same exponents, and the 
numbers with exponents above unity must be identical. In 
many cases this is impossible, at which times the operation can 
only be indicated. 

Example. Add 3 X 10^ to 56 X lo^ 

Solution. 3 X 10^ = 3,000 X 10^. This can obviously be 
added to 56 X 10®; the sum is 3,056 X 10^. 

Example. Subtract 56 X 10^ from 3 X 10^. 

Solution. Reducing as above, we obtain (3,000 X 10*) — 
(56 X 10^) = 2,944 X 10^ as the remainder. 

Practically these operations of addition and subtraction are 
restricted to notation in powers of ten; they are not much used 
and the above illustrations will serve as the rule for doing them. 

Factoring Exponents. An exponent can be factored, one 
of its factors with the number going inside a parenthesis and 



14 ELEMENTARY ELECTRICAL CALCULATIONS 

the other factor going outside of the parenthesis. Or if a quan- 
tity is thus expressed with a factored exponent, the parenthesis 
may be removed by multiplying the factors for a new exponent. 

Example. Factor the exponent of lo^^. 

Solution. io33 = do")' or (10^)11. 

Example. Express the square root of 11^ with factored 
exponents. 

Solution. Vii^ = III = (ii»)i 

Example. In the practical system of electro-magnetic 
units the unit of mass is taken as io~" grams, that of length 

as 10^ centimeters, and that of time as i second. The C.G.S. 

1 1 
unit of potential is M^L^T-'^, that of resistance is LT-^. Cal- 
culate the value of the practical unit of current and of resist- 
ance in C.G.S. imits. 

Solution. By Ohm's law the current is equal to potential 

divided by resistance, or — ^^ — . To effect the indicated 

division subtract the exponents of similar quantities in the 
denominator from those of similar quantities in the numerator. 
Calling the practical unit of current / and performing the 
division, we have I = M^L^T-^. Substituting for M and L 
their values as used for the practical units, we have / = 
(iQ-^^)^ (10®)^ I. It is unnecessary to repeat M, L, and T in 
this formula, as each is equal to unity. Thus if X 10® = i X 
10^ = 10®. Continuing the operation, (10^)^ X (10-")^ = 
lo^ X io~^^^ = io~^ = 10-^ = iV. This has to be multiplied 
by i-^ As the quantities are dissimilar the exponents cannot 
be added; the indicated operation may be performed. io~^ X 
i-^ = iQ-^ -7- I = Tu. Therefore the ampere is one-tenth of a 
C.G.S. unit of current. 
By Ohm's law resistance is equal to the potential divided by 

the current, or — ; — : = L T-^. This result is obtained by 



EXPONENTIAL NOTATION 15 

the process of exponential division, namely subtracting alge- 
braically the exponents of identical quantities. Substituting 
for L its value, 10^, and remembering that i has the same value 
whatever its exponent, we have, for the numerical value of 
resistance, 10^ X i~^ == 10^ C.G.S. units. 

These operations are given to illustrate the application and 
operations with powers of ten and with fractional exponents. 
The significance will be developed later, when they may be 
recurred to with advantage. 



CHAPTER III. 
MECHANICS AND PHYSICS. 

Units, Simple and Compound. — Significance of Multiplication and 
Division of Different Kinds of Units. — Centimeter-gram-second 
System. — Mass. — Weight and Gravitation. — Space. — Rate 
Units. — Velocity. — Acceleration. — Force. — Dyne. — Gravity. — 
Weight. — Acceleration of Gravitation. — Energy. — Available 
Energy. — Entropy. — Potential Energy. — Kinetic Energy. — 
Varieties of Energy. — The Erg. — Power or Activity. — British 
System of Units. — Value of Kinetic Energy. — Equivalence of Units. 
— The Volt-Coulomb. — The Joule. — The Watt. — Heat Energy. — 
Relations of Different Units. — Relation of Power Units and Energy 
Units. — Efl&ciency. — Central or Radiant Force. — Force of a Plane 
on a Point near its Surface. — Theory of Dimensions. — Dimensions 
of Mechanical Units. 

Units, Simple and Compound. Units are the basis of 
measurement. Thus a foot or an inch is a unit of length; a 
gram or a pound is a unit of weight. For different things there 
are different units ; electric current, mechanical energy and heat, 
for example, have each their own units. 

The same thing may be measured in some cases by different 
systems of units. Thus length may be measured by the English 
system, the inch, the foot, and other units being employed, or by 
the metric system, the centimeter, the meter, and other units 
being employed, or by any one of the many systems used by 
different nations at different epochs. There are two systems 
of electric units — the electro-magnetic and the electro-static 
systems. Most calculations in electric engineering are in elec- 
tro-magnetic units. 

A simple unit is one into which only one factor enters. The 
centimeter and the pound are examples of simple units. 

A compound, unit is composed of two or more simple units. 
The metric system unit of velocity is a compound unit; it is 
one centimeter per second. A horse-power in the English 

i6 



MECHANCIS AND PHYSICS 1/ 

system is expressed as the power which can raise 33,000 pounds 
one foot in a minute. This quantity is expressed by a com- 
pound unit ; it is 550 foot-pounds per second and three simple 
units make it up, the foot, the pound, and the second. 

A compound unit, expressed by units with hyphens between 
them, implies multiplication of one by the other. 

Example. 7,200 foot-pounds of energy are expended in 
lifting a weight 109 feet. What is the weight ? 

Solution. The hyphen indicates that in the compound unit 
" foot-pounds " the feet are multiplied by the pounds. There- 
fore 109 feet must be multiplied by such a number of pounds 
as will give a product of 7,200. This number is 66.06, which is 
the number of pounds lifted. The product of 109 feet by 
66.06 pounds gives 7,200 foot-pounds. 

When the words " in a," the word " per," " in," or " a " 
stand between units, the division of the coefficients of the units 
into one another is implied. These are rate units. 

Example. A trolley car is timed over a distance of 10 
rails; each rail is 60 feet long. It covers the distance in 15 
seconds. How many feet in a second does it travel ? 

Solution. The car travels 600 feet per 15 seconds. The 
rate per second is obtained by dividing the coefficient of feet, 
which is 600, by the coefficient of seconds, which is 15. 600 -r- 
15 =40, the number of feet per second. 

Significance of Multiplication and Division of Different 
Kinds of Units. The constituent units of a compound unit, 
written with a hyphen, practically never have any coefficient of 
either of the constituent units expressed, unity being the coeffi- 
cient. A gram-centimeter is one gram multiplied by one centi- 
meter. 10 gram-centimeters are ten times the above. The 
coefficient, 10, applies to and multiplies the compound unit, 
gram-centimeter. 

In the use of units the meanings of the words " multiplication " 



1 8 ELEMENTARY ELECTRICAL CALCULATIONS 

and ** division " are extended so as to include the practical 
multiplication and division of different classes of units with 
each other. Thus a foot-pound is taken as a multiplication of 
a foot by a pound. 150 foot-pounds is taken as the multi- 
plication of 10 pounds by 15 feet, or as the multiplication of 
15 pounds by 10 feet, all of which are impossibilities. 

The difficulty may be met thus. Suppose 10 feet are to be 
multiplied by 15 pounds. 10 and 15 may be treated as coeffi- 
cients of feet and pounds, and the multiplication may be expressed 
as 10 X 15 = 150 compound units, each compound unit being 
a foot-pound. 

Centimeter-Gram-Second System. The scientific bases 
of units are the centimeter, gram, and second. On these 
a whole series of units electrical, mechanical, and others have 
been founded. The system is called the C.G.S. system. 

The centimeter is the unit of length. It is approximately 
four-tenths of an inch, or one-thirtieth of a foot. 

The gram is the unit of mass. It is the quantity of matter 
contained in a gram. Its relation to weight is abstracted from 
its status as a unit of mass. A gram weighs approximately 
fifteen grains (15.4324-). 

The second is the unit of time. 

Mass. Mass is quantity of matter. In a given portion of 
matter it is invariable, and is unaffected by the relations of the 
portion of matter to the rest of the universe. The mass of a 
gram would be the same at any place, — at the center of the 
earth, on the surface of the earth, on the surface of a planet, 
or elsewhere. 

Weight and Gravitation. Weight is mass acted on by 
gravity. The weight of a given portion of matter depends upon 
the intensity of the force of gravity. As this force varies in 
intensity at different parts of the universe, the weight of a given 
portion of matter varies also. In the center of the earth there 



MECHANICS AND PHYSICS 1 9 

would be no weight. On the surface of the planet Jupiter the 
weight of the mass of a gram or other quantity would be much 
greater than on the earth. Weight is greater at the poles of the 
earth than at the equator; a pound would weigh less at the 
equator than in the polar regions, principally on account of 
centrifugal force, partly on account of the difference between 
the equatorial and polar diameters of the earth. 

The force of gravity of the earth is called its gravitation. 
As the earth is indefinitely large with respect to the masses cf 
engineering, the gram, pound, ton, and others, it acts upon such 
masses w^ith a force proportional to their masses. It there- 
fore tends to impart to any such mass the same velocity after 
acting on it for a second. This velocity is about 981 centi- 
meters per second and varies with the latitude. 

Space. Space may be of one, two, or three dimensions. 
Space of one dimension is length, that of two dimensions is 
area, and that of three dimensions is volume. Generally when 
space is spoken of in this book it is space of one dimension or 
length that is referred to. 

Rate Units. One class of units expresses rate. If a cur- 
rent of electricity flows at the rate of i coulomb per second, 
it flows at the rate of i ampere. If electrical energy is expended 
or developed at the rate of 10 megergs (see page 24) per 
second, the unit expressing this rate is the watt. 

Example. A generator delivers 8,050 coulombs in an hour. 
What is the rate ? 

Solution. There are 3,600 seconds in an hour. Dividing 
the total amount of electricity delivered by the time in seconds 
gives the rate per second. 8,050 -7- 3,600 = 2.24 amperes. 

Example. What quantity of electricity will be delivered by 
an amperage of 13 in 5 minutes ? 

Solution. 13 amperes of current is a rate of 13 coulombs 
in a second. In 5 minutes there are 300 seconds. Therefore 



20 ELEMENTARY ELECTRICAL CALCULATIONS 

the current will pass 13 X 300 = 3,900 coulombs in 5 
minutes. 

Sometimes a unit simple in form is so used as to imply rate. 
Thus when a velocity of 2 or any other number of length units 
is spoken of, what is meant is a rate of so many of the units 
per second. 

Velocity. Velocity is lineal space traversed per second. 
A velocity of 10 feet is 10 feet traversed per second. 

Example. A train of 60-foot cars is timed as it passes an 
observer. It requires 4 seconds for 6 cars to pass. Calculate 
the velocity. 

Solution. 6 cars each 60 feet long have a total length of 360 
feet. The space traversed in a second is equal to 360 -r- 4 = 
90 feet. "The velocity of the train is 90 feet. 

Acceleration. Change of velocity per second, which is the 
rate of change of velocity, is acceleration. If it is an increase of 
velocity it is positive ; if it is a decrease of velocity it is negative. 
The word " positive " is not generally expressed but is under- 
stood ; the word " negative " must be expressed when the 
acceleration is of that sign. 

Example. A body falls in a vacuum, starting from rest. At 
the end of 5 seconds it has a velocity of 4,905 centimeters. What 
is the acceleration, it being understood that the change in velocity 
has been constant and uniform ? 

Solution. The change of velocity per second is found by 
dividing the total change by the time required. This is 

4,905 -^ 5 = 981. 
The acceleration is 981 centimeters. 

Example. A ball is thrown into the air. It starts with a 
velocity of 96.6 feet and ceases to rise after 3 seconds. What 
is its acceleration ? 

Solution. Proceeding as before, we find 
96.6 -^ 3= 32.2. 



MECHANICS AND PHYSICS 21 

The acceleration is numerically 32.2 feet; but as it is a decreas- 
ing acceleration it is negative, or — 32.2. 

The last three problems refer to rate units, so that they are 
solved by division of the first by the second quantity. The 
quantities are respectively 360 feet per or in 4 seconds, 4,905 
centimeters in 5 seconds, and 96 feet in 3 seconds. 

The C.G.S. units of velocity and acceleration are i centi- 
meter per second. 

Force. Force is that which can change the state of motion 
or rest of a mass by acting on it for a period of time. It is that 
which can impart velocity to a mass by acting on it for a period 
of time. 

. The Dyne. The C.G.S. unit of force is the dyne. It is the 
force which in one second can produce an acceleration of one 
centimeter in a mass of one gram ; the force which can impart 
unit velocity in unit time to unit mass. The acceleration 
acquired in a second by a mass of a gram is the measure of 
the force acting on it, it being assumed that it is perfectly free to 
move and that its inertia is the only resistance it opposes to 
motion. 

Example. A mass of 1,107 grams has a velocity of 1,471 
centimeters imparted to it in i^ seconds. How many dynes have 
acted on it ? 

Solution. The acceleration is 1 ,47 1 -^ ij = 980I centimeters. 
If it were one gram the force would be 980! dynes. As it is 
1,107 grams, the force is 1,107 X 980} = 1,085,598 dynes. 

The word "force" is by usage sometimes applied inaccurately. 
Electro-motive force is not really a form of force and cannot be 
measured by force units. 

Gravity. Gravity is a true force. The gravity of the earth 
can impart in one second a velocity of 981 (about) centimeters to 
a mass. In other words it acts upon a gram with a force of 981 
dynes. As the earth's gravity acts upon masses in proportion 



22 ELEMENTARY ELECTRICAL CALCULATIONS 

to their masses, it can impart this velocity to a mass of any 
dimension. 

Weight. The action of the earth's gravity on mass is weight. 
A gram is a measure of weight just as it is a measure of mass. 
Force is often expressed in weight units, such as the pound, 
gram, or ton. Weight units of force are inexact, because the 
attraction of the earth varies with locahty. 

Example. How many dynes does a kilogram weight repre- 
sent ? 

Solution. A kilogram is equal to i,ooo grams. The force of 
gravity acts upon a gram with a force of 981 dynes. Therefore it 
acts upon a kilogram with a force of 1,000 X 981 = 98i,ooodynes. 

As there are i ,000 milligrams in a gram, a dyne is approximately 

represented by ■^-— = 1.02 mg. (milligrams). 
90 1 

Acceleration of Gravitation. The force of gravity is 
measured by the acceleration it can impart to masses of matter. 
The gravitation of the earth varies from 978.1 to 983 centimeters, 
which is its acceleration and which can be expressed in any 
other unit of length, as in feet or inches. Thus the acceleration 
of terrestrial gravity is about 32.2 feet. 

There are therefore two kinds of force units ; one kind is based 
on inertia and in the C.G.S. system is the dyne; the other is 
based on the earth's gravitation and in the C.G.S. system is the 
gram. One class is that of inertia units; the other is that of 
gravity units. 

Gravity acts on unit mass with a force numerically equal to 
the acceleration it imparts, which as we have seen is the velocity 
imparted in a second, about 981 centimeters. 

To reduce inertia units of force to gravity units divide by the 
acceleration of gravitation; in the C.G.S. system divide by 981. 
To reduce gravity units to inertia units multiply by the same 
figure. 



MECHANICS AND PHYSICS 23 

Example. What force is required to overcome the inertia of 
a mass of 1,000 kilograms so as to impart to it a velocity of 
9 kilometers per hour in 5 seconds ? 

Solution. 1,000 kilograms = 1,000,000 grams. 9 kilograms 
per hour =900,000 centimeters per 3,600 seconds =250 centi- 
meters per second. This velocity is acquired in 5 seconds, 
giving an acceleration of 250 -7- 5 =50 centimeters. Multiply- 
ing the acceleration by the mass it is imparted to gives the dynes 
of force, or 1,000,000 X 50 = 50,000,000 dynes. Dividing this 
by the acceleration gives the gravity units of force, or 50,000,000 
-^ 981 = 50,968 grams of force. 

Energy. Energy is the overcoming or capability of over- 
coming a resisting force along a lineal space traversed. It is 
the product of force by space traversed by the point of applica- 
tion of the force. It is therefore measured in two kinds of 
units, inertia and gravity units, just as force is measured, and 
the reduction of one to the other is effected by division or mul- 
tiplication by the acceleration of gravitation exactly as in the 
case of force units. 

Energy may be expended or developed ; neither operation can 
occur alone; both must be simultaneous and equal in amount. 
If a given amount of energy is expended, an exactly equal 
amount of energy is developed. This is the doctrine of the con- 
servation of energy. 

Available Energy. — Entropy. Man's economic needs are 
served by transformation of energy, generally by the trans- 
formation of higher grade energy into lower grade. If all 
energy were of the same grade none would be economically 
available for the uses of mankind. Available energy is called 
entropy. The available energy of the universe is constantly 
diminishing. 

Coal and the oxygen of the air existing separately represent 
high grade energy. When coal is burned the two combine^ 



24 ELEMENTARY ELECTRICAL CALCULATIONS 

producing heat energy. A small part may be utilized by man. 
After burning the grade of the remaining and unutilized energy 
is so low that it cannot be used. Entropy has been lost, but 
energy remains unchanged. 

Potential Energy. Potential energy is the power of 
exerting energy, which may be present in inert matter owing to 
circumstances of position or other states. A weight raised to a 
height has the capability of exerting energy in its descent to its 
original position. This capability is potential energy. The 
separate existence of coal and oxygen represents potential 
energy. In burning the two combine; the potential energy is 
converted into active energy, which is heat in this case. 

Kinetic Energy. Energy due to motion is called kinetic 
energy. A cannon ball in motion can exert energy, as in 
piercing an armor plate, which energy is due to its motion and 
is kinetic energy. 

Varieties of Energy. There are many kinds of energy, 
such as mechanical, heat, chemical, and electric energy. 

The Erg. The C.G.S. unit of energy is the erg. It is the 
energy exerted by a force of one dyne acting along a path one 
centimeter long and acting in the direction of its motion. The 
point of application of the force moves along the path. It is 
an inertia unit. One million ergs are a megerg. 

The erg is an inconveniently small unit in many cases, and in 
such values as the above the megerg is used. To convert ergs 
into megergs, move the decimal point 6 places towards the 
left. 

Example. How many ergs are exerted in raising 15 kilo- 
grams a distance of 3 meters ? 

Solution. 15 kilograms = 15,000 grams. 3 meters = 300 
centimeters. Multiplying these gives the energy in gravity 
units; 15,000 X 300 = 4,500,000 gram-centimeters. Multi- 
pl3dng this by the acceleration of gravitation gives the equiva- 



MECHANICS AND PHYSICS 25 

lent in inertia units, or ergs. 4,500,000 X 981 = 4,414,500,000 
ergs = 4,414-5 megergs. 

Power or Activity. Rate of energy is termed power or 
activity. Power units can be reduced to ergs per second or to 
other energy units per second. Power or activity is the rate of 
expending, developing, or transforming energy. 

Example. In the last example what was the power 
exerted if the 15 kilograms was raised the 3 meters in 6 
seconds ? 

Solution. Since power is ergs per second in the C.G.S. 
system, to get the answer in that system for inertia units 
divide the energy by the time. 4,414.5 h- 6 =735.75 megergs 
per second. 

Example. 27 dynes act along a path 27 centimeters long. 
What is the energy in ergs ? 

Solution. 27 X 27 = 729 ergs. 

Example. If 729 ergs are exerted in overcoming a resisting 
force along a space traversed of 25 centimeters' length, what 
is the force ? 

Solution. 729 -7- 25 = 29.16 dynes. 

British System of Units. The British system of units is 
founded on three fundamental units, the foot, the pound, and 
the second. 

The unit of velocity is a rate of i foot per second. 

The unit of acceleration is the velocity acquired in i second. 
It is obtained in any given case by dividing the velocity acquired 
by the time required to attain it. 

Example. In 13 seconds a car attains a velocity of 7 feet 
per second. What is its acceleration ? 

Solution. It is 7 -r- 13 = 0.538 foot. 

The inertia unit of force in this system is the poundal. It is 
the force which can impart a velocity of i foot per second to 
I pound by acting on it for i second . 



26 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. If in the last example the car weighed 40,000 
pounds what force was exerted on it in poundals ? 

Solution. As a velocity of 0.538 foot was imparted to 
40,000 pounds in i second, the poundals of force were 0.538 X 
40,000 = 21,520 poundals. 

The gravity unit of force is the pound avoirdupois. Gravity as 
a force produces an acceleration of about 32.2 feet per second 
(which corresponds to and is equal to 981 centimeters). 
Therefore all gravity units of the British system may be reduced 
to inertia units by multiplying by 32.2. 

Example. How many poundals in 6 pounds weight ? 

Solution. 6X32.2=193.2 poundals. 

Inertia units of the British system are reduced to gravity 
units by dividing by the same figure. 

The inertia unit of energy is the foot-poundal ; it is the force of 
a poundal exerted along a path of i foot. The gravity unit is 
the foot-pound, which is the energy required to raise i pound a 
height of I foot. 

The British unit of power is the horse-power. It is the rate 
of energy required to exert 550 foot-pounds of energy per second, 
or 33,000 foot-pounds per minute. 

Example. Returning to the example on page 25, what was 
the force exerted in pounds, and what energy was expended in 
10 feet of the car's progress ? 

Solution. As the poundals are in the English system, use the 
acceleration of gravity in feet, 32.2. Then 2 1,5 20 -^ 3 2. 2 =668. 3 
pounds. The energy is force multiplied by space traversed, 
or 21,520 X 10 = 215,200 foot-pound als, or 668.3 X 10 = 6,683 
foot-pounds. 

Example. How many foot-poundals are exerted by a 150- 
pound man in going up a flight of stairs 10 feet high ? 

Solution. 150 pounds X 10 feet = 1,500 foot-pounds, and 
i>5oo X 32.2 = 48,300 foot-poundals. 



I 



MECHANICS AND PHYSICS 2/ 

Value of Kinetic Energy. If a force a acts upon a body 
free to move, it will impart to it a uniformly increasing velocity, 

and at the end of a time t the velocity will be - , / indicating 

lineal space. This is acquired velocity. The average velocity 

will be one-half of this, or -X - * The space traversed will be 
2 t 

the product of the average velocity by the time required to trav- 
erse it ; this product is — X — X ^ = — • The value of the 

2 t 2 

force acting on the body is the product of the acceleration it 
imparts to the body multiplied by the mass, which we will call m. 
Acceleration is the quotient of velocity divided by the time 

required to attain it, or — '- t = -^ » ^^^ ^^^ force is -j- • 
t I I 

Energy is equal to the product of force by the space along which 

it is exerted; this product is then -r-X — = — :r = ~" w^^- 

t^ 2 2 r 2 

The above formula gives the value of kinetic energy in inertia 
units, such as ergs or foot-poundals. To reduce these to gravity 
units the value given by the formula must be divided by the 
acceleration of gravity in the system to which the formula has 
been applied. 

Example. Calculate the energy in a mass of ii grams 
moving at the rate of 210 centimeters per second. 

Solution. By the formula we have energy = J X n X 210^ 
= 242,550 ergs, which divided by the acceleration of gravity 
in the metric unit, 981 centimeters, gives the value in the 
gravity system as 247 centimeter-grams, or 247 grams raised to a 
height of I centimeter against the attraction of gravity. 

Example. What is the energy in a 150-pound projectile 
moving at the rate of 2,000 feet per second ? 

Solution, i X 150 X 2,000^ = 3 X lo^ foot-poundals, or 
dividing by 32.2 = 9,316,770 foot-pounds. 



28 ELEMENTARY ELECTRICAL CALCULATIONS 

Equivalence of Units. The equivalents given below are 
used to change from one system to the other. 

I foot = 30.48 centimeters and i centimeter =.0328 foot. 

I inch = 2.54 centimeters and i centimeter = .3937 inch. 

I pound = 453.6 grams and i gram = .0022 pound or 15.432 
grains. 

From the above data : 

Example. Calculate the value of the poundal in dynes. 

Solution. 30.48 (centimeters) X 453-6 (grams) = 13,826 dynes. 

Example. Calculate the value of the pound in dynes. 

Solution. 453.6 (grams) X 981 (centimeters) = 4.45 X 10^ 
dynes. 

Example. Calculate the value of the foot-poundal in ergs. 

Solution. 453.6 X 30.48 X 30.48 = 421,408+ ergs. To find 
the number of ergs in a foot-pound multiply 421,408 ergs X 
32.2 = 13,569,338= 13-57 megergs. 

Example. How many ergs are there in a horse-power acting 
for one second ? 

Solution. Take 13.57 megergs as the value of the foot- 
pound. Then we have i horse-power-second = 550 foot- 
pounds = 550 X 13-57 = 7,464 megergs = 7,464,000,000 ergs. 

Example. What energy in ergs is exerted in raising 80 
pounds to a height of 5 feet ? 

Solution. 80 X 5 = 400 foot-pounds. Using the factor or 
equivalent determined in previous problem we have 
13,569,338 X 400 = 5,427,735,200 ergs. 

The Volt-Coulomb. The practical unit of electric energy 
is the volt-coulomb. This is based upon the centimeter, gram, 
and second, being a C.G.S. system unit. It is equal to 10^ 
C.G.S. units of energy, which units are ergs. It is a watt-second, 
and is equal to a joule. 

Example. Calculate the equivalent of a volt-coulomb in 
foot-pounds. 



MECHANICS AND PHYSICS 29 

Solution. The volt-coulomb is equal to 10^ ergs, which are 
10 megergs. The foot-pound is equal to 13.57 megergs. 

Therefore a volt-coulomb is equal to = 0.737 foot-pound. 

13.57 
The Joule. Another unit of energy is the joule. It is 

equal to 10^ ergs, 10 megergs, and therefore to the volt-cou- 
lomb, and in many cases can be used as a synonym of the volt- 
coulomb. 

The Watt. The practical unit of electric power is the volt- 
ampere, which is equal to i volt-coulomb per second, which is 
equal to i joule per second. This unit is called the watt. 
As the volt is equal to 10^ C.G.S. units, and as the ampere is 
equal to iq-^ C.G.S. units, it follows that the volt-ampere or 
watt is equal to 10^ X 10-^ = 10^ C.G.S. units of power, which 
C.G.S. unit is i erg per second. The watt is equal to 10^ ergs 
or I joule per second. 

Example. What is the value in watts of 10" ergs per minute ? 

Solution. 10" -i- 60 = 1,666 X 10® ergs per second. Divid- 
ing this by 10^ gives the watts of power. 

1,666 X 10® -T- 10^ = 166.7 watts. 

Heat Energy. Heat is a form of energy. As it is due to 
motion of the particles of matter, it is kinetic energy. The 
C.G.S. unit of heat energy is the erg. It is sometimes called 
the primary unit of heat. 

Units of Heat Energy. The engineering and working 
units of heat energy are based on the heat required to impart a 
definite increment of temperature to a definite weight of water. 
The increment or increase of temperature is defined by refer- 
ence to one of the thermometric scales. As these are arbitrary, 
such units have an arbitrary relation to the C.G.S. system. 
Their values are determined by experiment and are not abso- 
lutely accurate. 

The heat required to raise the temperature of a gram of water 



30 ELEMENTARY ELECTRICAL CALCULATIONS 

1° C. is the therm, gram-degree, minor calorie, or simply calorie. 
It is sometimes called the secondary unit of heat. It is equal 
to 41.66 megergs, often given as 42 megergs. 1000 calories is 
the heat required to raise the temperature of a kilogram of 
water i°C. This is the kilogram-degree or major calorie. It 
is equal to 41,666 megergs. 

The heat required to raise the temperature of a pound of 
water i°F. is the British thermal unit, often written B.T.U. 
It is equal to 778 foot-pounds, approximately. 

Example. Calculate the value of the calorie in volt-cou- 
lombs. 

Solution. I volt-coulomb = 10 megergs. i calorie =41.66 
megergs. Therefore i calorie =4.166 volt-coulombs. 

Example. Calculate the value of the B.T.U. in volt-cou- 
lombs. 

Solution. I volt-coulomb = 0.735 foot-pound, i B.T.U. 
= 778 foot-pounds. Therefore i B.T.U. = 778 -f- 0.735 = 
1,058 volt-coulombs. 

Energy Units and Equivalents. Energy equivalents are 
approximate once the C.G.S. system is departed from. Thus 
the value of the B.T.U. is variously given, ranging from 772 
to 778 foot-pounds. The difficulties attending the accurate 
determination of this equivalent account for the discrepancies. 

Relations of Different Units. The scope and meaning of 
equivalents of units may be illustrated by the watt-second. 
This unit of electric energy is equal in value (a) to lo^ ergs, 
(b) to I joule, (c) to 0.24 calorie, (d) to 10,193 gram-centimeters, 
(e) to 0.00095 B.T.U., (/) to 0.737 foot-pound, and to others. 

The watt-second is a compound unit, for which the joule is 
often a synonym, and is equal to a rate of energy of one watt 
exerted for one second. Following up the equivalents given 
above we see that it can (a) and (b) impart to a gram a 
velocity of 10^ centimeters; (c) it can heat 0.24 gram of water 



MECHANICS AND PHYSICS 3 1 

I® C; {d) it can raise 10,193 grams to a height of i centimeter; 
(e) it can heat 0.00095 pound water i°F.; (/) it can raise 0.737 
pound to a height of i foot. 

The relation of the C.G.S. units to each other is purely decimal. 
The relations of the so-called engineering units to each other 
are fixed and arbitrary; they are arbitrary in the sense that the 
values of the engineering units were not based on any system 
of uniform relationship. 

Example. If 18 ounces of water are raised 5°F. in tem- 
perature, how many foot-pounds of energy will be absorbed ? 

Solution. As there are 16 ounces in a pound, 18 ounces = 
I.I 25 pounds. Multiplying this weight by the number of degrees 
through which it is raised gives the B.T.U.'s developed. 1.125 
X 5 = 5.625 B.T.U.'s. Applying the equivalent, 778, gives 
5.625 X 778 = 4,376.25 foot-pounds. 

Relation of Power Units and Energy Units. A power 
unit followed by a time unit with a hyphen intervening gives 
energy. Thus a horse-power-second is equal to 550 foot-pounds 
of energy. A watt-second is equal to a volt-coulomb, a unit of 
electric energy. Sometimes the hyphen is omitted. 

Example. Calculate the energy in 29 horse-power-minutes. 

Solution. There are 60 seconds in a minute. Therefore 
29 horse-power-minutes are equal to 29 X 60 = 1,740 horse- 
power-seconds. A horse-power-second is equal to 550 foot- 
pounds. 1,740 horse-power-seconds = 1,740 X 550 = 957,000 
foot-pounds of energy. 

Example. Calculate the electric and mechanical energy in 
5,040 watts acting for 20 minutes. 

Solution. 20 minutes = 20 X 60 = 1,200 seconds. The 
energy of the above watt-minutes is 5,040 X 1,200 = 6,048,000 
watt-seconds, or volt-coulombs of electric energy. The volt- 
coulomb is equal to the joule, so we may read for the volt- 
coulombs 6,048,000 joules of mechanical energy. 



32 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. A man runs up a flight of stairs 12 feet high in 
5 seconds. He weighs 150 pounds. What power does he 
exert in horse-power and in watts ? 

Solution. In 5 seconds he expends 150 X 12 =1,800 foot- 
pounds of energy; the rate per second is therefore 1,800 -j- 5 
= 360 foot-pounds, which is equal to 360 -f- 550 = 0.655 
horse-power, because a horse-power is a rate per second of 550 
foot-pounds. Take the foot-pound as equal to 13.57 niegergs. 
Multiplying this by 360 gives 360 X 13.57 = 4j885.2 megergs, 
or 488.5 watts, because the watt is equal to 10 megergs, or to 
10^ ergs. 

Example. Taking a pound as 453.6 grams, and the foot as 
30.48 centimeters, calculate the equivalent of the horse-power 
in watts using 981 as acceleration of gravity. 

Solution. A horse-power is 550 foot-pounds per second. 
I foot-pound is equal to 30.48 X 453-6 = 13,825.728 gram- 
centimeters. Taking the acceleration of gravity as 981 
centimeters, the last figure multiplied by 981 gives dyne-centi- 
meters, or 13,825.728 X 981 = 13,563,039 ergs, because an 
erg is a dyne-centimeter. A watt is 10 megergs per second; 
a foot-pound, therefore, in watts is given by dividing the value 
in ergs by 10^, giving 1.356 watts. The horse-power is equal to 
550 foot-pounds, or in watts to 1.356 X 550 = 746 watts. 

This problem could be solved by simply multiplying 550 
(foot-pounds per second) by 13.57 and then dividing by 10. 

Efficiency. The relation of useful to total energy in any 
process is called efficiency. Useless energy is that which is ex- 
pended in overcoming friction and hurtful resistances generally. 
Useless energy is always developed at the same time with use- 
ful, and is manifested in the heating of bearing surfaces, of 
electric conductors, and in other ways. 

Efficiency is the ratio of the part of the energy utiHzed to the 
total energy expended. It is generally stated as a percentage. 



MECHANICS AND PHYSICS 33 

Example. 7 megergs are expended each second in driving 
a dynamo. 6 megergs per second are delivered to the system 
to be there utilized. What is the efficiency ? 

Solution. 6-^7= 0.857 is the efficiency decimally ex- 
pressed. It is 85.7 per cent. The remainder, 14.3 per cent, 
is wasted. 

Central or Radiant Force. If two points at a distance 
from each other attract or repel each other, the force exerted 
upon each one will vary inversely as the square of the distance. 

Example. Assume that two magnet poles attract each 
other with a force of 9 dynes when 2^ centimeters apart. The 
distance is increased to 3 centimeters. What will the attrac- 
tion be at this distance ? 

Solution. The proportion of the inverse squares of the dis- 
tances is 3 : 2J : : g : X = 5.0625 dynes. 

The attraction could have been expressed in other units, 
such as grains or grams. 

The attraction or repulsion exerted by two points upon each 
other varies with the product of the force of one body by that of 
the other. 

Example. Assume that there are two electrically charged 
pith-balls at such a distance from each other that they have 
with respect to each other 2 units of force and 3 units of force 
respectively. With what force will they attract each other ? 

Solution. The product of their individual attractions is 
3X2=6 units of force. The force may be measured in any 
convenient unit, such as dynes. 

Distance and the forces exerted by both points enter into the 
problem simultaneously in many cases. 

Example. One north magnet pole has 5 megergs force as 
regards its action on the south pole of another magnet situated 
3 centimeters distant from it. The south pole of the other 
magnet has 7 megergs force under identical conditions. Cal- 



34 ELEMENTARY ELECTRICAL CALCULATIONS 

culate their mutual attraction at the stated distance of 3 centi- 
meters and at 4 centimeters. 

Solution. Their mutual attraction is equal to the product of 
their individual attractions, or 

5 X 7 = 35 megergs. 

This is at the distance of 3 centimeters. For their attraction 
at the distance of 4 centimeters we must apply the rule of the 
inverse squares, thus: 

4^ : 3^ — 35 : ^ = 19.7 megergs. 

Example. At i centimeter distance the individual forces of 
two magnet poles are 9 and 17 dynes respectively. Calculate 
the combined attraction at the distance of 19 centimeters. 

Solution. The mutual attraction of the poles at the stated 
distance of i centimeter is 

9 X 17 = 153 dynes. 

The attraction at the distance of 19 centimeters may be 
obtained by the rule of the inverse squares. As the one term 
of the proportion is i, or unity, it is sufficient to divide the 
attraction at i centimeter by the square of the other distance, 

* ^^* 153 -J- 192 = 0.424 dyne. 

For the above rules to hold the poles or other things acting on 
each other must be small compared to the distance. The law 
then applies with reasonable accuracy. To be rigorously 
accurate the points acting on each other must be infinitely 
small. 

Force of a Plane on a Point near its Surface. A plane 
of indefinitely large size exerting force on a point does so with 
a force numerically expressed as 2 ntrm, in which o- (the Greek 
letter sigma) indicates the force per unit area of the plane and 
m the force of the point. The point might be a magnet pole 
and the plane the face of a magnet, or it might be a mass acted 
on by gravity. The force is the same irrespective of distance, 



MECHANICS AND PHYSICS 35 

provided the area of the plane is large enough. The law is 
deduced by calculus. 

Theory of Dimensions. By the use of what is known as 
the theory of dimensions the relations of mechanical and phys- 
ical quantities to each other are expressed in algebraic expres- 
sions into which only three quantities enter, namely, space, 
time, and mass. They were an invention of Fourier, and Clerk 
Maxwell brought them into prominence. They are not of 
frequent use in engineering work, but are of great value in the 
theoretical aspect, and as they are really very simple, should be 
studied. 

Exponential notation is used in the discussion of dimensions. 

Dimensions of Mechanical Units. Units in the abso- 
lute and in the practical systems of mechanical and electric 
units are derived directly from the three fundamental units of 
length, mass, and time, namely, the centimeter, the gram, and the 
second. These fundamental units are designated by the letters 
L, Mf and T. Mechanical units will first be treated. 

Velocity is equal to the length traversed in a given time 
divided by the time required to traverse the length in question. 
Its dimensions are L/T, or LT-^. 

Acceleration is the rate of change of velocity and is equal 
to the velocity acquired in a given time divided by the time 
required to attain such velocity. Its dimensions are therefore 
velocity divided by time, L/T-r- T, or LT-^, 

Example. A trolley car starts from rest and in 12 seconds 
is moving at the rate of 200 feet in 21 seconds. What is the 
velocity attained and what is the acceleration ? 

Solution. Substituting for L and T their values gives 

Velocity = LT-^ = 200 -i- 21 = 9.524 feet, or 290 cm. 
Acceleration = LT-^ XT-^ = 9.52 -^ 12 = 0.7936 feet, or 
24.19 cm. 
Force is measured by the acceleration it can impart per unit 



36 ELEMENTARY ELECTRICAL CALCULATIONS 

of mass. Its dimension is the product of mass by acceleration. 
It is clear that it takes more force to give acceleration of a given 
amount as the mass is larger. It takes twice the force to impart 
a given velocity to a mass of two grams that it does for a mass 
of one gram. Hence the multiplication is needed to express 
the force required to give acceleration to a mass, and the prod- 
uct of the multiplication is the number of force units required. 
Its dimension is the product of mass M by acceleration L T-^ ; 
M X LT-^ = MLT-"^ are the dimensions of force. 

Example. How many dynes were required to impart the 
velocity of the last problem, assuming the car to weigh 40,000 
pounds ? 

Solution. The mass of the car is 40,000 X 453-6 = 18,144 
X 10^ grams. The acceleration is 24.19 cm. Multiplying 
mass by acceleration gives 18,144 X 10^ X 24.19 =439 X 10® 
(about) dynes. 

This can be done by use of T and T^ for the last example. 
Force = MLT-'T-"^ = 18,144 X lo^ X 6,096 4- (21 X 12) = 
439 X 10^ (about) dynes. 

The factor 6,096 is the equivalent of 200 feet in centimeters. 

It will be observed that in this example there are two values of 
r, which have been indicated by T and T\ In other words, to 
use the dimensions directly it is necessary to take cognizance of 
the two values, as is done in the solution. Where two values of 
the same unit occur it is often possible to modify the dimensional 
formula so as to substitute for the square of L, M, or ^S" the 
product or rectangle of two such symbols, distinguished by prime 
marks or other distinguishing characters. 

Example. Apply the method as above to the first example. 

Solution. As there are two values of T, the formula can be 
modified to read L T-^ T^-^. Substituting for the symbols their 
values as given in the statement of the problem, we have 

Acceleration = 2oo-^(2IXI2)= — =0.7936 feet, as before. 

252 



MECHANICS AND PHYSICS 37 

This treatment is merely given as a process by which dimen- 
sional formulas can be directly applied to the solution of problems. 
Strictly speaking they should be regarded as expressing propor- 
tional relations of quantities. 

If a force acts along a path it must necessarily overcome a 
resistance and its point of application moves along. The 
exercise of force along a path, or along a length traversed, is 
energy. Its dimensions are the product of force MLT-^ by 
space L, which gives ML^T-^. As constant factors do not 
enter into dimensions, a factor, J, which is required to make 
this expression numerically correct in calculations, is omitted. 
In the numerical sense the expression for energy is J ML^T-^. 
By using the rectangle of factors instead of their squares, calcu- 
lations can be made by simple substitution. 

Example. A force acts upon a mass of 200 grams and in 
5 seconds moves the mass a distance of 750 cm. What energy 
is expended upon the mass ? It is understood that the mass 
derives all this motion from the force and that the force is uniform. 

Solution. Let L represent the distance traversed by the 
body and let L' represent the distance it would traverse in 
5 seconds at the velocity imparted by the force. U is equal to 
2 L, because for the purposes of the problem the mass may be 
taken as starting from rest, and it receives a perfectly uniform 
increase of velocity. The distance traversed in the 5 seconds, 
which is 750 cm., is equal to the average of the starting and 
terminal speeds, referred to 5 seconds. The starting speed 
is o ; the terminal speed is V ; the average is (o + L') -^ 2 = Ly2 
= 750; whence U = 1,500. Substituting in the formula, 
energy = ML X L' -^ T^ gives 

Energy = (200 X 750 X 1,500) -^ 5^ = 9,000,000 ergs. 

Power is the rate of expenditure of energy. Its dimensions 
are the quotient of energy ML^T-^ divided by time Ty giving 
MVT-K 



38 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. What power was expended upon the mass in the 
last example ? 

Solution. The 9 X lo® ergs of energy were expended in T 
seconds, the power was 9 X lo*^ -7- 5 = 18 X 10^ C.G.S. units of 
power. As a watt is 10^ C.G.S. units, the power is 0.18 watt. 

Momentum is the quantity of motion in a body, as often 
defined. It is mass M multiplied by the velocity L/T at 
which it moves, giving MLT-^. It is momentum to which 
Newton's laws of motion refer. 

PROBLEMS. 

Reduce 60 feet in 3 seconds to feet per second. 

Ans. 20 feet per second. 

Reduce 5,280 feet per minute to feet per second, 

Ans. 88 feet per second. 

A train runs at 64 miles an hour; how many feet per second is 
its speed? Ans. 93.9 feet per second. 

A trolley car is timed over a distance of ten rails ; each rail is 60 
feet long, and it covers the distance in 16 seconds. Calculate the 
velocity. Ans. 37.5 feet. 

A machine exerts 7,000,000 foot-pounds of energy per hour. 
Calculate the horse-power. Ans. 3.53 + horse-power. 

A train moves at the rate of 11 car lengths in 9 seconds ; each car 
is 60 feet long ; how many C.G.S. units of velocity does this reduce 
to? Assume the foot to be 30.48 cm. in length. 

Ans. 2,235 C.G.S. units. 

What velocity is 310 feet in 59 seconds? Ans. 5.25 feet. 

A car starts from rest and in 120 feet attains a speed of 10 miles 
per hour. Calculate the acceleration. Ans. 0.89 feet. 

A mass of 21 grams has a velocity of 390 cm. imparted to it. 
A period of 7 seconds was required to impart this velocity. How 
many C.G.S. units of force were exerted? Ans. 1,170 dynes. 

How many ergs are there in a mass of 5 grams, free to fall 
I meter, the acceleration of gravitation being 981 cm. ? 

Ans. 490,500 ergs of potential energy. 



MECHANICS AND PHYSICS 39 

How many ergs are present in a mass of 397 grams moving at a 
speed of 2,340 cm. per minute? 

Ans. 301,918.5 ergs of kinetic energy. 

A mass of 5 grams has a rate of motion of 3 cm. per 6 seconds 
imparted to it in a period of 5 seconds. Calculate the (a) final 
velocity, (b) acceleration, (c) force, (d) energy present at the end 
of the period. Ans. (a) 0.5 cm. 

(6) 0.1 cm. 

(c) 0.5 dyne. 

(d) 0.625 erg. 

Calculate the foot-pounds of energy in a 20-ton car moving at 
the velocity of 30 feet. Ans. 559,006 

If brought to rest in 11 seconds, what power is exercised in 
doing this? Ans. 50,819 foot-pound-seconds. 

92.4 horse-power. 

A i2-kilowatt generator runs for 35 minutes. What are the joules? 
Ans. 252 X 10 ^joules, or 25,200,000 joules. 

How many foot-poundals are there present in a 20-ton car mov- 
ing at 26 miles an hour? Ans. 29,083,022 foot-poundals. 

If the above car is brought to rest in 30 seconds what horse- 
power is expended in the braking? Ans. 54.73 horse-power. 

A generator produces a current of 50 amperes at 112 volts. Cal- 
culate the horse-power. Ans. 7.51 horse-power. 

How many iio-volt lamps averaging one-half ampere each are 
to be allowed to the horse-power? Ans. 13.56 lamps. 

If the earth and the moon were each of twice its present diameter 
and of the same specific gravity, how would the attraction of one 
for the other be affected ? 

Ans. As mass varies with the cube of like linear parts the mass 
of each would be 8 times as great as now, and the attraction would 
be 8 X 8 = 64 times the present. 

A magnet pole of 2 dynes strength is 30 cm. from one of 3 dynes 
strength. Calculate the attraction of one for the other. 

Ans. 0.0066 dyne. 

To what height must 51 grams be raised to give 392 gram-centi- 
meters? • Ans. 7.686 cm. 



40 ELEMENTARY ELECTRICAL CALCULATIONS 

If a dynamo armature is making 750 revolutions per minute, 
how many revolutions does it make per second ? Ans. 12.5. 

One ampere at a pressure of one volt is a volt-ampere. Express 
in correct form 11 amperes at 12 volts pressure. 

Ans. II X 12 = 132 volt-amperes (not 11 volts X 12 amperes 
= 132 volt -amperes). 

What is the rate of electric energy if a generator delivers 36,000 
megergs in an hour? Ans. 10 megergs per second = i watt. 

A piece of some material weighs 251 pounds on a spring balance 
at the equator. What will it weigh at the pole, taking the ratio of 
the gravitation at the two points as 983 and 978? 

Ans. 252.3 pounds. 

How many calories does a 150-kilowatt generator develop per 
hour? Ans. 1,296 X 10 ^ 

How many joules are there in a kilowatt-hour? 

Ans. 36 X 10'. 

Express the last two answers in ordinary numeration. 

Ans. 129,600,000 calories. 
3,600,000 joules. 

550 watts are passed through the coil of an electric boiling appa- 
ratus. The boiler contains one pint of water. After 6 minutes the 
water is brought from 60° F. to boiHng (212° F.). What is the 
efficiency? Ans. 81 per cent, 

(i pint of water = i pound; 1,058 watt-seconds = i B.T.U.) 

If a dynamo produces energy at the rate of 3,500 B.T.U. per 
hour, what is the rate of output? Ans. 1,025 watts. 

A motor absorbs 500 watts. It drives a pump and raises in an 
hour 1,500 gallons of water 90 feet. What is the efficiency of the 
system ? Ans. 45 per cent, (nearly) 



CHAPTER IV. 

OHM'S LAW. 

Three Factors of an Active Circuit. — Ohm's Law. — Several Appliances 
in One Circuit. — Application of Ohm's Law to Portions of a Circuit. 
— Simple Method of Expressing Ohm's Law. — Proportional Form 
of Ohm's Law. — Fall of Potential. — RI Drop. — Counter Electro- 
motive Force. — Problems. 

Three Factors of an Active Circuit. In an active circuit 
there are three factors on which its action depends. These are 
current, electro-motive force, and resistance. The relation of 
these three in a circuit through which a current is passing is 
embodied in Ohm's law. 

Ohm*s Law. It has been found by experiment that in a con- 
ductor of given resistance the intensity of current varies with the 
electro-motive force (abbreviated as e.m.f.). It has also been 
found that in conductors of different resistances the currents 
due to the same e.m.f. vary inversely as the resistances. When 
both e.m.f. and resistance vary, the relation of the current to such 
variations is expressed in the proportion given below, in which 
E, I, and R indicate e.m.f., current intensity, and resistance 
respectively. 

Let a unit current be assumed to pass through a resistance R. 
For this to take place there will have to be an e.m.f. Now 
assume that another current of intensity I is to pass through the 
same resistance instead of the current i, then the e.m.f. will have 
to be different, and the relation of these quantities will be given 
in the proportion i - I • • R ■ E 

which is a proportion expressing Ohm's law. 

From the last proportion are derived the three equations or 

formulas r E 

/=|; E^RI; R^f. 

41 



42 ELEMENTARY ELECTRICAL CALCULATIONS 

These are the most used expressions of Ohm's law. 

In words the law is generally expressed thus : 

The current intensity is equal to the electro-motive force 
divided by the resistance. 

The electro-motive force is equal to the product of the resist- 
ance by the current. 

The resistance is equal to the electro-motive force divided by 
the current. 

Example. An electro-motive force of 5 volts is expended on 
forcing a current through a resistance of 10 ohms. What is the 
intensity of the current ? 

Solution. Applying the first form of the law, we divide the 
e.m.f. by the resistance, obtaining 

Intensity of current = 5 volts -7- 10 ohms = J ampere. 

Example. What e.m.f. is required to force 10 amperes 
through a resistance of 105 ohms ? 

Solution. Here the second form of the law may be employed. 
Multiplying the resistance by the current gives the e.m.f., or 
105 X 10 = 1,050 volts. 

Example. An e.m.f. of 27 volts forces a current of 5 amperes 
through a wire. What is the resistance of the wire ? 

Solution. The third form of the law applies here. Dividing 
the e.m.f. by the current intensity we have 

27-^5=51 ohms. 

The different classes of problem of which the above are 
examples can all be done by the one form of Ohm's law. It is 
more convenient to apply the three forms respectively as illus- 
trated in the three examples. 

Several Appliances in One Circuit. The elements which 
determine the . intensity of a current are the total resistance 
and the total e.m.f. in the circuit, irrespective of the distribu- 
tion of the same therein. Assume several batteries to be dis- 



I 



OHM'S LAW 43 

tributed on a circuit. Each battery introduces two things in 
the system, e.m.f. and resistance, both of which have to be taken 
into consideration. 

Example. An electric circuit has distributed along its course 
three sources of e.m.f. A battery of 8 ohms resistance and 2 
volts e.m.f. is connected by a wire of 107 ohms resistance to a 
second battery of 20 ohms resistance and 21 volts e.m.f. This 
is connected by a wire of i ohm resistance to a third battery of 
5 ohms resistance and 10 volts e.m.f. The circuit is closed by 
a wire of 17 ohms resistance, connecting the first to the third 
battery. All the batteries work together, being of identical 
polarity. Calculate the current. 

Solution. The total resistance of the circuit is made up of 
the resistance of the batteries added to that of the conductors. 
This gives 

jR = 8 + 107 + 20 + I + 5 + 17 = 158 ohms. 
The total e.m.f. is the sum of the e.m.f. 's of the batteries. 
This gives £ = 2 + 21 + 10 = 33 volts. 

By Ohm's law we have 

33 volts 
/ = -^ — -. — = 0.200 ampere. 
158 ohms 

Application of Ohm's Law to Portions of a Circuit. 

Ohm's law applies to any portion of a conductor through which 
a current is passing. The intensity of the current passing 
through a circuit is equal not only to the e.m.f. of the circuit 
divided by the resistance thereof but is also equal to the e.m.f. 
expended on any portion of the circuit divided by the resist- 
ances of the same portion. This will be found treated in one 
of its bearings elsewhere in this book. 

Let e, e^, . . . e^ denote the e.m.f. 's expended on portions of 
an electric circuit, and let r, r^, . . . r^ denote the resistances 



44 ELEMENTARY ELECTRICAL CALCULATIONS 

of the respective portions corresponding thereto; then, calling 
the intensity of current /, as before, we have 

I = — ■ — *— , and also / = -=-, and so on. 

r -\- r^ -]-... + Tn r r^ 

This formula expresses an almost obvious law, that while the 
e.mi. expended on portions of an active circuit varies with the 
resistances of the same portions, the current is uniform in all 
parts of the circuit. 

Example. A voltmeter connected to two parts of an active 
circuit shows 40 volts. The resistance of the wire included 
between the points where the voltmeter terminals are connected 
is jf ohm. What current is passing through the circuit ? 

Solution. We have as the values of e and r of the formula 

last given 40 and |f , so that 

T 16 8s 1 

/=4o-^— =-^ = 42* amperes. 
17 2 

Simple Method of Expressing Ohm's Law. A very 

ingenious way of representing and of memorizing Ohm's law is 
embodied in the following device : 

E 
RXl' 

If any one of the elements is removed , the relative position of 
the other two gives the value of the third in terms of the other 
two. Thus, if from the group we remove E, R X I are left ; 
therefore the value of E in terms of R and I is R X 1 , or the 
product of R and / as it should be. If jR is removed from the 
group, E/I remains, giving the value of R in terms of E and /, 
which is E divided by /. In the same way, if I is removed from 
the group its value remains, just as in the other cases, namely, 
E/R, or E divided by R. 

Example. The e.m.f. between the ends of a conductor is to 
be determined. Its resistance is 15 ohms and a current of 
5 amperes is maintained through it. Apply the diagram. 



OHM'S LAW 45 

Solution. Removing E from the diagram leaves R X I. Sub- 
stituting for R and / their values gives the value of E or of the 
e.m.f. as 15 X 5 = 75 volts. 

Example. Take the e.m.f. between the ends of a conductor 
as 30 volts and its resistance as 15 ohms. What current will 
pass through it ? 

Solution. Removing the symbol of current / from the 
diagram there remains E/R^ and substituting the values of resist- 
ance and e.m.f. from the statement of the problem we obtain 
30/15 = 2 amperes. 

Example. Let a current of 25 amperes be maintained by 
an e.m.f. of 29 volts. What is the resistance of the conductor 
in which it is so maintained ? 

Solution. Removing R from the diagram leaves E/l, and sub- 
stituting as before for E and / their values from the conditions 
as stated in the problem we find 

R =— = — = 1.16 ohms. 
^ 25 

Proportional Form of Ohm's Law. Ohm's law can be 
employed in the proportional form. It is a form but little 
employed in practical work but is available for practice by the 
student. Three principal forms may be stated, corresponding 
to three principal forms of the law. 

The resistance varies with the quotient of the electro-motive 
force divided by the current. 

F K 
R, : R, : : f : f- (i) 

The current varies with the quotient of the electro-motive 
force divided by the resistance. 

F E 
^^■'^■- R,- R, (=) 

The electro-motive force varies with the product of the resist- 
ance by the current. E^ : E^ : : RJ^ : R^I^. (3) 



46 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. Compare the resistance of two conductors with 
potential differences between their ends of 7 and 16 volts 
respectively and currents of 13 and 17 amperes respectively 
produced thereby. 

Solution. Applying formula (i) we find 

^1 • ^2 : : ^ • 7: = 0-539 ■ 0-94I. 

This proportion gives more than the mere ratio, for the third 
and fourth terms of the proportion give the values of R^ and 2^ 
namely, 0.539 ohm and 0.941 ohm respectively. 

Example. The e.m.f. of a battery is 10,7 volts and its 
resistance is 50 ohms. A conductor of 1,101 ohms resistance 
connects the terminals of the battery. Compare the current 
with that which would be produced through the same con- 
ductor by a battery of 6.42 volts and of 30 ohms. 

Solution. Proportion (2) gives 

T T 10.7 . 6.42 

7i : I2 • • ^ • — ^— = 0.0093 ' 0.0057. 

1,151 1,131 

Again we have the result expressing not only the proportion 
but also giving the actual figures, namely, 0.0093 ampere and 
0.0057 ampere. 

The denominators of the two fractions are the sum of the 
resistances of the batteries and of the external resistances in 
each of the two cases. 

Example. Compare the e.m.f.'s producing in two con- 
ductors of tV and 29 ohms resistance respectively currents of 
200 and I.I amperes. 

Solution. By the third proportion we find 

E^ : £2 • • 200 X — : 29 X i.i = 10.52 : 31.9, 

which as before gives the actual values as well as the ratio of 
the e.m.f,'s. 



I 



OHM'S LAW 47 

Fall of Potential. The e.m.f. of a generator, such as a 
battery or dynamo, is the potential difference maintainable 
between its terminals when no current is being taken from it. 
When a circuit is open the potential difference between its ends 
is equal to that at the terminals of the battery or generator. 
When a circuit is closed and active the potential difference 
between any two points upon it depends upon the current main- 
tained in it. 

The potential difference existing between two points of a 
circuit is often called drop of potential, potential drop, fall of 
potential, voltage, and the like. 

By Ohm's law the potential drop in any part of an active cir- 
cuit is equal to the resistance of that portion multiplied by the 
current. Calling the potential drop in a portion of the circuit 
£', and calling the resistance of the same portion R\ we have 

E' = R'L 

The current is the same in all parts of a circuit, hence the 
current is indicated by the letter / without any mark. 

JJJ Drop. The fall of potential in a portion of an active 
circuit as just explained and as expressed by this formula is 
very often called the " RI drop." The RI drop is the fall of 
potential in any part of an active circuit the resistance of which 
part is R. As it applies to any part of a circuit it logically 
includes the drop in the whole circuit. 

Example. A circuit is passing a current of lo amperes. 
What is the RI drop in a portion of resistance if ohms ? 

Solution. Applying the formula, we multiply the current by 
the resistance, 

i?X/ = ioXif= — = 14.3 volts. 

Example. Calculate the RI drop which will exist in a 
resistance of 1 1 ohms added in series to a circuit of 5 ohms and 
7 volts. 



48 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. When ii ohms are added to the circuit the resist- 
ance of the circuit will beiiH-5 = i6 ohms. The current will 
be 7 -^ i6 = 0.4875 ampere. The RI drop of the ii-ohm por- 
tion of the circuit will be, by the formula, 11 X0.4875 = 5.2625 
volts. 

Example. What must the resistance of a 20-volt circuit be 
if an additional resistance of 5 ohms introduced in circuit has 
an RI drop of 3 volts ? 

Solution. Applying the formula, we have 

£ = 7? X /, or 3 = 5 X /, 

and therefore the current in the circuit after the extra resistance 
is introduced is j_, ^^^^^^^ 

The same formula will give the resistance of the whole circuit 
after the additional resistance is introduced. Here E is 20 volts 
and / is I ampere. Substituting gives 

20 = 7? X f and 2? = 20 X I = ZZ-ZZ ohms. 

The original resistance of the circuit was therefore ZZ-ZZ ~ 5 
= 28.33 ohms. This is the answer to the problem. 

Example. What must the e.m.f. of a circuit be if an addi- 
tional resistance of 7 ohms develops within itself an RI drop of 
9 volts, the original resistance of the circuit being 12 ohms ? 

Solution. With the new resistance the total resistance of the 
circuit will be 12 + 7=19 ohms. The current in the new 
part of the circuit can be calculated from the formula as before : 
E = RI, or 9 = 7 X /, and I = j amperes. 

The e.m.f. of the circuit is given by the same formula : 

E =RI, or E = igX- = — = 24.43 volts. 

7 7 

Example. The total resistance of a circuit is 29.37 ohms. 
The e.m.f. of the generator is 17 volts. Calculate the RI drop 
of the whole circuit. 



OHM'S LAW 49 

Solution. By Ohm's law the current is equal to 17 -7- 29.37 
= 0.58 amperes. Multiplying the current by the resistance 
as before we have for the RI drop 29.37 X 0.58 = 17 volts. 
The RI drop of the circuit is equal to the e.m.f . of the generator, 
as it should be. 

Counter Electro-motive Force. If generators, such as 
batteries or dynamos, are put in series with each other and are 
of opposite polarities, which means that they tend to produce 
currents in opposite directions, the e.m.f. produced by the com- 
bination will be equal to the difference of the sum of the e.m.f. 's 
of the generators working in one direction and the sum of the 
e.m.f.'s of the generators working in the other direction. 

If positive signs are assigned to e.m.f. of one polarity and 
negative signs to the other, the resulting e.m.f. will be equal to 
the algebraic sum of the e.m.f.'s. Generators thus placed are in 
opposition to each other. The e.m.f. of one generator tends to 
produce a current in the opposite direction to that which the 
generator in opposition tends to produce. A pair of battery cells 
connected in opposition would have the zinc plates connected, 
and the terminals from the other plates connected to the two 
terminals of the line. 

Example. Two cells of battery are connected in series and 
in opposition. One cell has 1.05 volts e.m.f.; the other has 
1.79 volts. What is the net e.m.f.? 

Solution. Subtracting the e.m.f. of one battery from that 
of the other gives net e.m.f. = 1.79 — 1.05 = 0.74 volt. 

Example. Four cells A, B, C, and D are in series with 
each other. Their e.m.f.'s are respectively i volt, ij volts, 2 
volts, and 1.66 volts. The zinc plate of A is connected to 
the zinc plate of 5, the carbon plate of B to the carbon plate of 
C, and the zinc plate of C to the zinc plate of D. The 
circuit is completed by a wire connecting the carbon plate of D 
to the carbon plate of A. What is the e.m.f ? 



50 ELEMENTARY ELECTRICAL CALCULATIONS 



Solution. Cells A and C are of the same polarity; cells 
B and D are of opposite polarity to that of A and C, but are of 
the one polarity necessarily with each other. Call the polarity 
of B and D positive and that of A and C negative. The e.m.f. 
of E and D then is ij + 1.66 = 3 volts positive. The e.m.f. 
of A and C is 1 + 2=3 volts negative. The net e.m.f. is 
equal to the difference of the positive and negative e.m.f. 's, 
namely, 3 — 3 = o, or zero. 

The conditions of the problem are illustrated in the diagram 



H-l— l-h 



A B 



C D 



Example. Two batteries are placed in opposition to each 
other. One has a resistance of 2.14 ohms and a voltage of 
1. 01. The other has a resistance of 5 ohms and a voltage of 
2.07. They are connected in series in a circuit of 209 ohms 
external resistance. Calculate the current. 

Solution. The total resistance is 2.14 + 5 + 209 = 216.14 
ohms. As the batteries are in opposition, the e.m.f. of the two 
is found by subtracting the e.m.f. of one from that of the other. 
This gives 

2.07 — 1. 01 = 1.06 volts. 

Applying Ohm's law we have 
1.06 



I = 



216.14 



0.00497 ampere. 



PROBLEMS. 

The same e.m.f. acts upon circuits of 10 and of 3 ohms resistance. 
What is the proportion of the currents resulting? 

Ans. i/io : 1/3 = i : ^-33, or the current in the 3-ohm circuit will 
be 3.;^^ times as strong as in the lo-ohm circuit. 



OHM'S LAW 5 1 

The same e.m.f. can produce currents of ii and of 5 amperes 
through different circuits. What are the relative resistances of the 
circuits? 

Ans. i/ii : 1/5 == I : 2.2, or the current in the 5-ohm circuit is 
2.2 times as strong as that in the 11 -ohm circuit. 

What is the relative strength of current produced through identical 
resistances by 5 and by 7 volts? Ans. 5 : 7 = i : 1.4. 

What will be the relative resistances of circuits in which the above 
e.m.f.'s will produce currents of identical intensity? Ans. i : 1.4. 

The e.m.f. expended on a portion of an active circuit is 23 volts and 
the resistance of this portion is /^ ohm. What e.m.f. would be 
expended on a portion of the circuit of i| ohms resistance? 

Ans. 200.1 volts. 

A current is passing through two incandescent lamps in series. 
One lamp has a resistance of 107 ohms, the other has a resistance of 
200 ohms. The e.m.f. expended on the lamp of higher voltage is 
found by voltmeter to be 73 volts. What e.m.f. is expended on the 
other? Ans. 39.055 volts. 

A circuit has a total resistance of 3 ohms and includes a generator 
of 2 volts e.m.f. Into this circuit a generator of i volt in opposition 
is introduced whose resistance is 3 ohms. What effects are produced ? 

Ans. New voltage, i. 

New resistance, 6 ohms. 
Original current, § ampere. 
New current, I ampere. 

A generator has 2 ohms resistance and 18 volts e.m.f. It sends a 
current of 2 amperes through a wire of 4 ohms resistance and also 
through a battery of i volt e.m.f. placed in opposition to the generator. 
What is the resistance of the battery? Ans. 3^ ohms. 

A storage battery of 20 volts and 0.33 ohms has been producing a 
current of 15 amperes through a circuit. Another battery of 20 cells, 
each cell of 1.08 volts, can produce a current of 2.7 amperes through 
the same circuit. If the two batteries are connected in opposition 
what will the data of the circuit be ? 

Ans. Resistance, 8.33 ohms. 

Electro-motive force, 1.6 volts. 
Current, 0.192 amperes. 



52 ELEMENTARY ELECTRICAL CALCULATIONS 

A generator sends lo amperes through 25 ohms. It can send only 
9 amperes through 29 ohms. What is its internal resistance ? 

Ans. II ohms. 
This problem can be done by the use of the following equations: 

= 10 (i); ■ =9. (2) 



X + 2^ .T + 29 

There are two storage batteries of identical constants E and r. A 
single one produces a current of 2 amperes in a circuit of 30 ohms. 
If the two batteries are connected in parallel the current becomes 2.2 
amperes. What are the constants of the batteries ? 

Ans. 73.3 volts. 
6.6 ohms. 

The above example can be done by the use of the following 

equations: 20 + r = ■ — (i); 30 H — = — (2); 

20 2 22 

in which E and r are the unknown quantities. 

A battery has an e.m.f. of 3.21 volts. The potential difference 
between its terminals within the battery when they are connected by 
a wire of 8 ohms resistance is 1.07 volts. What is the resistance of the 
battery? Ans. 16 ohms. 

The above example can be done by algebra by the use of the 
following equation: 

n 21 I 0*7 

■^ — = —~-^ , in which x is the resistance of the battery. 
8+^8' ^ 

The RI drop in a conductor is 57.25 volts; the current is 0.75 ampere; 
what is its resistance ? Ans. 76.33 ohms. 



CHAPTER V. 

RESISTANCE. 

Linear Conductors. — Resistance and Weight of Linear Conductors. — 
Parallel Conductors. — Distribution of Current in Parallel Conductors. 
— Resistances of Conductors in Parallel. — Combined Resistance 
of Two Conductors. — Combined Resistance of any Number of 
Conductors. — Specific Resistance. — Circular Mils. — Effect of 
Temperature of Conductors on their Resistance. — Problems. 

Linear Conductors. A linear conductor is one of uniform 
cross section. An ordinary wire is an example. 

The resistance of linear conductors of identical material varies 
directly with the length and inversely with the cross-sectional 
area. Consequently it varies inversely with the square of 
identical elements of the cross section. Thus in the case of 
circular-section conductors, such as wires, the resistance varies 
inversely with the square of the diameter or the square of the 
radius. 

This gives the following proportions, in which l,R, A, d, and r 
indicate the length, resistance, area, diameter, and radius 
respectively of conductors, subscript numbers serving to indi- 
cate individual conductors. 



' ' ' A A, ' ' d' ' (d,y • V {ry 



A :A, : : d" : ^)2 . . ^2 . (^^)2 



L .k. 
R ' R. 



(I) 

(2) 



I .1^ \ '.RA \ R,A, : : Rd" : T^W,)^ : : Rr'' : R,{r,)\ (3) 



Example. Compare the resistance of two conductors whose 
lengths are 341 and 361 feet respectively and whose cross- 
sectional areas are 27 and 37 mils respectively. 

53 



54 ELEMENTARY ELECTRICAL CALCULATIONS 
Solution. Substituting these values in formula (i) gives 

K : K : : — : — = 120 : 07, 

27 37 

which gives the relative resistance of the two conductors. The 
conductor R^ has y^g- the resistance of the conductor R. 

Example. Compare the resistances of two wires, one of 
length 3 and diameter 2, the other of length 6 and diameter 8. 
Call diameter d. 

Solution. — Here formula (i) is applicable, substituting for 
areas of cross sections the squares of Hke cross-sectional ele- 
ments. In this case it is the square of the diameter which is to 
be used. This gives 

7?.p/..i.i._l.^_3..A_« 

(P d^ 2^ 8^ 4 64 

The resistance of R is eight times that of i?'. 

Example. — A wire is to be installed in place of one 0.357 
inch in diameter, the new wire to be of one-half the resistance 
of the original one. Calculate its diameter. 

Solution. Applying proportion (2), using the square of the 
diameter in place of the cross-sectional area, we have 

^ : ^' : : ^2 . j^2 . . (3^7)2 ; ^2 . . I . i = i : 2, 

^ 2 

or x^ must be twice as large as the square of the original wire, 
which is (357)^ = 127,449. Twice this, or 254,898, is the square 
of the diameter of the new wire; its diameter is \/2 54,898 = 
504 mils. 

Example. There are two wires of relative resistances 57 
and 91 and of relative diameters 31 and 79 respectively. The 
first wire is 673 feet long. How long is the other? 



RESISTANCE 



55 



Solution. — Applying proportion (3), substituting the square 
of the diameters for the cross-sectional areas, we have 

673 : ^ : • 57 X (31)' : 91 X (79)' = 54,777 ' 567,931; 
therefore by the rule of ratio and proportion 

^ ^ 673 X 567,931 ^ 382,217,563 ^ 6 f^^^^ 
54,777 54,777 

Example. 1,000 feet of circular wire have a resistance of 
105.6 ohms. What will the resistance i? of a square wire of 
the same diameter be ? 

Solution. From geometry we know that the area of a circle 
is 0.7854 times the area of the square of the same diameter. 
Proportion (i) is applicable. As the lengths of the two wires 
are the same the proportion is simplified by substituting for / in 
the numerator of the fraction unity, or i. This gives 



R : 105.6 : 



and R = 105.6 X 0.7854 = 82.9 ohms. 



I 0.7854 

Example. A rectangular conductor has a cross section of 
} inch by 2 inches and is 20 feet long. Compare its resistance 
with that of a round bar 2 inches in diameter and 21 feet long. 
Solution. The cross-sectional area of the rectangular bar is 
I X 2 = li square inches. The cross-sectional area of the 
circular bar is 3.14 square inches. Applying proportion (i) and 
calling the resistance of the rectangular bar i, we have 

20 21 
li 3-14 

2T 20 



whence 



X = 



= 0.500, 



3-14 li 

or the rectangular bar has one-half the resistance of the round 
one. 

The proportions on page 53 can be expressed as equations 
thus: 



R = 



RMi 

AL 



(I); 



. -if «, 



' - g; "• 



56 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. A conductor of 117 feet in length and 980 square 
mils in cross-sectional area has a resistance of i ohm. What is 
the resistance of 1,987 feet of wire of 9,862 square mils cross- 
sectional area? 

Solution. — Substituting in formula (i) the values of the 
problem gives 

R = ^ ' — = 1.69 ohms. 

9,862 X 117 

Example. A conductor 75 feet long and 39 square mils, 
cross-sectional area is of 20 ohms resistance. What is the cross- 
sectional area of a conductor of 19 ohms resistance and 89 feet 
long, of the same alloy? 

Solution. Applying formula (2) we have 

. 39 X 89 X 20 . 

A = -^ = 48.7 square mils. 

19 X75 

Example. A conductor is 1,000 feet long, 150 square mils, in 
cross-sectional area, and has a resistance of 53 ohms. Another 
conductor is of 185 square mils cross-sectional area and of 75 
ohms resistance. What is its length? 

Solution. Applying formula (3) we have 

^^1,000X75X185^^ ^^^^^ 
53 X 150 

Resistance and Weight of Linear Conductors. The 
resistance of a wire may be referred to its weight and length. 
The weights of equal lengths of wire are in proportion to their 
cross-sectional areas and to the reciprocals of their resistances. 
If the relative resistances of two wires or other linear conductors 
and of known length and weight are to be calculated, the weights 
of equal lengths are found and the calculation is based on them 
as if they were cross-sectional areas. 



R : R 



^ ' ' wt. per foot ' wt.j per foot 



RESISTANCE 57 

Example. Two bars are ^;^ and 31 feet long. The first 
weighs 31 pounds, the second weighs 23 pounds. Calculate 
their relative resistances. 

Solution. One foot of the first bar weighs 31/33 = 0.939 
pounds, and one foot of the second weighs 23/31 = 0.742 pounds. 
Dividing the length of each bar by the weight of one foot as 
determined above gives the relative resistances, ^t, -t- 0.939 
= 35.14 and 31 -^ 0.742 =41.78 are the relative resistances. 
The proportion can then be written 

35.14 : 41.77 : : 1 : X = 1.19-, 

or, if we call the resistance of the 33-foot bar unity, the resistance 
of the 31-foot bar is 1.19— . 

Let / be the length of a conductor and w be its weight. Then 
its weight per unit of length is w/l. Its relative resistance is 
given, as we have seen by dividing the length by the weight per 
unit of area. This is 

/ WW 

The relative resistance of a conductor is expressed by the 
quotient of the length squared divided by the weight. 

The problem given above can be solved by this expression 
directly. Substituting the values of the weights and lengths of 
each conductor in the formula we have for the relative resistances 
iSSf -^ 31 = 35-1 and (31)2 --*" 23 = 41.8, and 351 : 41.8 : : 
I : 1. 1 9 as before. 

Example. There are two wires. 96 feet of one weighs 0.28 
pound, 118 feet of the other weighs 0.30 pound. 100 feet of 
the first have a resistance ot i ohm. How many feet of the other 
will have a resistance of i ohm? 

Solution. Applying the formula, the relative resistances are 

— ) = 96^ -i- 0.28 = 32,914 and 118^ -r- 0.30 = 46,413. 
w / 



SS ELEMENTARY ELECTRICAL CALCULATIONS 



As the lengths of linear conductors are directly proportional to 
their relative resistances, we have 

32,914 : 46,413 : : 100 : x = 141 feet. 

This rule is based upon the weight of equal lengths of the con- 
ductors. It applies to any lineal conductors, irrespective of the 
shape of their cross sections. 

Example. 3 feet of bus bar weigh 5.25 pounds. Calculate 
the resistance. 

Solution. Consulting a wire-resistance table, any wire may 
be selected as a standard. Thus 0000 wire weighs 639.6 pounds 
to the 1,000 feet, with a resistance of 0.051 ohm. Therefore 
3 feet of this wire weigh 0.6396 X 3 = 1.9188 pounds and have 
a resistance of 0.000,051 X 3= 0.000,153 ohm. The inverse 
proportion then holds: 

5.25 : 1. 9188 : : 0.000,153 : x = 0.000,056 ohm. 

Parallel Conductors. A group of conductors which start 
in one point and terminate in another are in parallel. This 

condition is shown in 
the figure. If a number 
of conductors are con- 
nected across the space 
between two leads of 
low resistance com- 
pared to that of the 
conductors, they are 
also in parallel. This 
condition is shown in 
the lower figure. 
The correct conception of parallel conductors is that the 
potential difference maintained between their ends is identical 
for all. 




RESISTANCE 59 

Distribution of Current in Parallel Conductors. The 

current passing through each of a set of parallel conductors is 
calculated by Ohm's law. It is therefore inversely propor- 
tional to the relative resistance of the conductor in question 
referred to that of the others. As conductance is the recipro- 
cal of resistance, if conductance is used in the calculation the 
proportion becomes a direct one. The general statement may 
be put thus: 

The current passing through single conductors of a set of 
parallel conductors varies directly with the reciprocal of the 
individual resistances. If the resistances of the members of 
a set of parallel conductors are denoted by a, h, c, . . . n, the 
reciprocals of the resistances will be denoted by i/a, i/b, i/c, 
. . . i/n. The sum of the reciprocals of all the conductors will 
be denoted by i/a + i/b + i/c . . . + i/n. The proportion 
expressing the law just given is the following: 

I , I I II 

a c n a 

Example. Assume that there are three conductors of re- 
sistances 2, 3, and 4 ohms respectively which are connected in 
parallel. If a current is passed through them it will be divided 
among them. The proportion of current which will pass 
through each of them is to be determined. 

Solution. Taking the reciprocals of the resistances and 
applying the proportion as above gives 

1 1 — :— ::i:a;, and 

2342 

12 + 8 + 6 12 12 . ^, , . 

: — : : I : jc = -— of the whole. 

24 24 26 

This gives the proportion of the current which will pass 
through the conductor of conductance, 1/2 i.e. of resistance, 
2 ohms. Proceeding in like manner for the other conductors, 



6o ELEMENTARY ELECTRICAL CALCULATIONS 

we find that the current through the conductor of resistance, 
3 ohms, is ^V a-iid that the current through the conductor of 
resistance, 4 ohms, is -i^ of the whole. 

By this method is found the relative current which will go 
through each of a set of parallel conductors. If the total cur- 
rent is known it is only necessary to multiply it by the respec- 
tive fractions to determine the current in amperes which will 
pass through each conductor. 

Example. In the case just calculated assume that a total 
current of 19 amperes passes through the conductors. It is 
then increased to 22 amperes. Calculate the current which 
will pass through each conductor in each case. 

Solution. For the first case proceed as follows: 

— X 19 = 8.77 amperes. — X 19 = 5.85 amperes. 
26 26 

6 

— X 19 = 4-39 amperes. 
20 

For the second case proceed in the same way. 

— X 22 = 10.15 amperes. — X 22 =6.77 amperes. 
26 26 

6 

-- X 22 = 5.08 amperes. 
26 

The wire of 2 ohms resistance will pass of the first current 
(19 amperes) 8.77 amperes; of the second current (22 amperes) 
10.15 amperes. The 3-ohms wire will pass of the same re- 
spective currents 5.85 amperes and 6.77 amperes. The third 
wire will pass of the same 4.39 and 5.08 amperes. 

As a test of the correctness of the operations add the cur- 
rents thus : 

8.77 -f 5.85 -}- 4.39 = 19 amperes and 
10.15 -I- 6.77 -f- 5.08 = 22 amperes. 



RESISTANCE 6 1 

The fractions of the total current adding up to the total 
current in each case goes to prove the correctness of the opera- 
tions. 

It is evident that it would have given the same result if the 
value of the total current in amperes had been used as the third 
term of each proportion instead of unity. This is the proper 
procedure when there is question of only one current. 

The proportion can also be expressed as a formula or equa- 
tion. Calling the total current / and the fractional currents 
7„, /j, Ic, ' ' ' Iji) 3-iid the resistances of the parallel conductors 
a,b,c, . . . n, as before, the proportion and the equation become 

-+f+---- +- :-::/ (total current) : 4. (i) 
a b c n a / a v / 

la (current in conductor a) 
I (total current) 



^(i + 1 + i... + i). (,) 

\a b c n) 



For currents in conductors 6, c, and any others substitute the 
resistance of the conductor whose current is to be calculated 
for that of a in the second term of the proportion (i/^, i/c, etc., 
instead of i/fi^). In equation (2) substitute 6, c, or what- 
ever the resistance of the particular wire may be for a in the 
fraction i/a in the second member of the equation, so that it 
will read thus : 

/j or /^ (or whatever it may be) = 



b ox c 



-.(i + i + i.-.i) 

\a b c n) 



The same may be expressed in the form of a rule. 

The current passing through one of a group of parallel con- 
ductors is equal to the total current divided by the resistance 
of the conductor, which quotient is again divided by the sum of 
the reciprocals of the resistances of the individual conductors. 



^'= 5 



62 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. A current of 27 amperes passes through three 
conductors in parallel of the respective resistances of 5, 7, and 
9 ohms. Calculate the current which will go through each one. 

Solution. Substituting in formula (2) gives 

^^/l + l + l^=l7 ^143 = ,, 

VS 7 9/ 5 315 
and consequently for the other two conductors 

^ ^ ^ ^ 143 ^ 8.5 ; I3 = -^ -^ ^3 = 6.6. 
7 315 9 315 

Adding the results together as a test of the correctness of the 
operations we obtain 

1 1.9 + 8.5 + 6.6 = 27, going to prove the correctness of the 
work. 

The wire of 5 ohms resistance passes 11.9 amperes, the wire 
of 7 ohms resistance passes 8.5 amperes, and the wire of 9 ohms 
resistance passes 6.6 amperes. 

The operation would be shortened by multiplying 27 by ^— ^> 

which gives 5.95. Then the resistance of each wire divided 
into this figure gives the current of that wire. Thus : 

59-5 -^ 5 = II-9; 59-5 "^ 7 = 8.5; 59.5 -^ 9 = 6.6, as before. 

Example. Four parallel conductors have resistances of J, 
if J, and i ohm respectively. A current of y ampere passes 
through them. Calculate the current passing through each one. 

Solution. Applying formula (2) gives 

/j = iZ7 ^(, + 3 + 7 + ,) = lZ7^i = ^. 

2 1/2 ^ ' ^ 21 2 147 

^ 21 3 147 
'^ 21 7 147 
^ 21 9 147 



' RESISTANCE 63 

Adding together these currents as a test of the calculation 

we obtain 

2,3,7,9 21 I 

1 — — -i — * — 1 — — — — = — ampere. 

147 147 147 147 147 7 

As the sum of the partial currents is equal to the total current 
the test goes to prove the correctness of the operation. 

Example. Assume the resistances of four parallel con- 
ductors to be §, 7, f , and | ohm respectively. Calculate the 
distribution of a current of f ampere among them. 

Solution. Proceeding as before we find 

7. =^^/l + L + i4-i')= ?Z7^ 826 _ i8o_^ 

2/3 \2 3 4 5/ 2/s 120 2,891 

In the same way we find 

r — 280 J. _ 150 ^ J. _ 216 

^"2,891 ' ^ "" 2,891 ' ^ " 2,891' 

The sum of these fractions is 

— -— = — ampere, as before, going to prove the correctness 
2,891 7 

of the process. 

2 
In this problem the operations reduce to multiplying — '- 

7 

— = — — = by the reciprocals of the resistances of 

120 5,782 2,891 

^u J ^ rru 120 ^^ 3 360 180 

the conductors. Thus — — - X — = -"^ = -7; — ampere cur- 
2,891 2 5,782 2,891 

rent, for the conductor of f ohm resistance, as before. The 
same method can be applied to the others. 

Resistance of Conductors in Parallel. If several con- 
ductors are placed in parallel their combined resistance will be 
less than that of any single one. The sign of combination of 
the resistances of parallel conductors is a semicolon placed 



64 ELEMENTARY ELECTRICAL CALCULATIONS 

between the figures or letters indicating the resistances of the 
respective conductors. 

Thus a ; h indicates the combined resistance of two parallel 
conductors, one of resistance a, the other of resistance Z>, when 
in parallel. The combined resistance is always less than the 
resistance of either one of the conductors. 

The power of conducting electricity is termed conductance. 
It is proportional to the cross-sectional area of a conductor. 
If therefore two conductors are in parallel with each other, the 
conductance of two is equal to the sum of their conductances. 

As the conductance of a conductor is equal to the reciprocal 
of its resistance, the conductance of several conductors in parallel 
with each other is equal to the sum of the reciprocals of their 
resistances. It follows that the reciprocal of this sum is equal 
to the resistance of the conductors in parallel. 

To calculate the resistance of conductors in parallel, add 
together the reciprocals of their resistances. The reciprocal 
of this sum will be the resistance of the conductors in parallel. 

Example. Combine the resistances 2; 4; 6 ohms. 

Solution. The conductances of the conductors in question 
are J, J, and J practical units (sometimes but very seldom 
called mhos). As they are in parallel their conductances must 
be added to obtain their aggregate conductance. 

L -I- 1 4- i. — 24+12+8 _ 44 ^ 1 1 
2 4 6 2X4X6 48 12 

As resistance is the reciprocal of conductance, the combined 
resistances 2 ; 4 ; 6 ohms are the reciprocal of this conduct- 
ance, xf or 1.0909 ohm. 

Example. Calculate the resistance of three conductors in 
parallel of J, J, and ^ ohm resistance respectively. 

Solution. The conductances of the conductors are 2, 4, and 
6 ohms. Their united conductance is 2 + 4+6 = 12 units. 
The resistance is the reciprocal of the conductance, or tV ohm. 



RESISTANCE 6$ 

Example. What is the resistance of 7, 9, and J ohms in 
parallel ? 

Solution. — \ f-2 =2^1 = — - = the conductance. The 

7 9 63 

resistance is the reciprocal of the conductance, which is — '— 

142 

= 0.444 ohm. 

Combined Resistance of Two Conductors. — Suppose two 

resistances are to be combined in parallel. Call them a and b. 

Their reciprocals are i/a and i/b. Adding these by the rule 

for fractions dves : , which is the conductance and whose 

reciprocal is the resistance. This is • 

^ a-\- b 

To calculate the resistance of two conductors in parallel, 
divide the product of their resistances by the sum. 

Example. Calculate the resistance of 45 and 61 ohms in 
parallel. 

Solution. By the rule just given it is (45 X 61) -^ (45 + 61) 
= 2,745 -4- 106 = 25.9 ohms. 

Example. Calculate the resistance of J and | ohm resist- 
ances in parallel. 

Solution. As before (LxA-^('- + A='-^^~=^ =^ 
\2 3/ V2 3/ 6 6 42 7 

= 0.285 ohm. 

The following may be given as a general arithmetical or 
algebraic law : 

To divide the product of two fractions by their sum, multiply 
their numerators together for a new numerator; for a new de- 
nominator multiply the numerator of one by the denominator 
of the other, and vice versa, and add the products. This rule 
gives an easy way of combining the resistances of two con- 
ductors in parallel. 

Example. Calculate the resistance of f and f ohm in parallel. 



^ ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. Applying the rule we have 

4.3.^^ IXJ - = 12. ohm. 

5 7 (4 X 7) + (3 X 5) 43 

Example. Combine the resistances f ; 3. 

Solution. An integral number can be expressed as a fraction 
having unity for its denominator and the number as its numer- 
ator; therefore 

4 . - _ 4 .1 

5 51 

and applying the rule gives 

4 . 3.^ ^_XJ = ilohm. 

5 I (4 X i) + 3 X 5) 19 

Combined Resistance of any Number of Conductors. 

The rule for the calculation of the resistance of any number of 
resistances in parallel may be thus deduced. Assume resist- 
ances fl, &, c, and d. Their reciprocals are i/(z, 1/6, i/c, and 
\ld. These are their conductances. Adding them gives the 
following expression : 

ahc -\- ahd -\- acd + hcd 
ahcd 

whose reciprocal is the combined resistance, which is 

ahcd 

abc -f- ahd + acd + hcd 

Expressed in words this may be thus put : 

To calculate the resistances of any number of resistances in 
parallel, multiply the resistances together for a numerator. 
For a denominator multiply together all possible combinations 
of one less than the given resistances and add the products. 
There will be as many such products as there are resistances. 
The result is the combined resistance. 



RESISTANCE 6/ 

Example. Calculate the resistances of the following resist- 
ances in parallel: i, 2, 3, and § ohm. 

Solution. By the rule given above we obtain the expres- 
sion 

1/2 X 2 X 3 X 2/3 

(1/2X2 X3) + (i/2X2X2/3) + (i/2X8X2/3) + (2X3X2/3) 

= -^ ohm. 

The operation of combining these resistances may be thus 
expressed : 

a ] b ; c ', d = 

^+^ + - + j 

abed 

It is obvious that when there are many conductors in parallel 
the calculation of their combined resistance by the general rule 
would be complicated and open to error in calculation. The 
following is a good method to use in doing the calculation, or 
as a check on the operation if done by the other method. 

Combine two resistances by the regular rule. Then com- 
bine the result with another of the resistances, and combine this 
result with yet another of the resistances, and so on until the 
last one has been combined. 

Example. Apply this method to the problem given above. 

Solution. The resistances are j ; 2 ; 3 ; I. Applying the rule 
we have 

i. _ i/2 X 2 ^2^ 2 ^ _ 2/.S X 3 ^ 6 ^ 6 2^ 6/17 X 2/3 
2'^ 1/2 + 2 5*5'^ 2/5 + 3 17' i7'3 6/17 + 2/3 

= ~ ohm. This is the same result as that reached by the 

13 

other method, and therefore operates as a check upon it. 
In many cases it is the better method of the two. 

Example. Combine the resistances 3; 5; 5; J; 9; If ohms in 
parallel. 



6S ELEMENTARY ELECTRICAL CALCULATIONS 

Solution, (a) 3; 5 = ^-^^ = -^ • W ^ ; 5 = '-^/^-^ 
' ' ^'^ 8 + 5 8 ^^ 8 '5 15/8 + 5 

= 75_. (^,)75 .1 _ 75/55 Xi/4 _ 75/220 ^21, 

55 55^4 75/55+1/4 3.000/220+55/220 355' 

y) ^; 9 = 75/355X9 ^ ^ . (,) .675.; IT. = 

355 75/355 + 9 3,270 3,270 18 

675/3,270 X 17/18 _ 11,475 _ 765 ^j^j^ 
675/3,270 + 17/18 67,740 4,516 

Specific Resistance. — Substances vary greatly in their 
power of conducting electricity. The resistance of a prism or 
cylinder varies directly with its length and inversely with its 
cross-sectional area. The resistance of a cyHnder or prism 
is therefore equal to , , 

(sp. r.)~^^. (i) 

section ^ 

In the above expression sp. r. is a coefficient whose value 
varies with the material of the conductor. It is called the 
specific resistance or resistivity of the material. 

The reciprocal of the above coefficient, i/sp.r., is the con- 
ductivity of the material. 

Specific resistance is the resistance between two faces of a 
cube of the material. In the C.G.S. system the unit cube is 
one centimeter cube, so that the C.G.S. unit of specific resist- 
ance is the resistance of a centimeter cube of the substance. 

Call p the specific resistance of any substance, and let R 
denote resistance. The dimensions of resistance are LT-^ (see 
page 122). The resistance of any lineal conductor is equal to 

-7 X ,0 = resistance = L T-^. 
A ' 

But for / can be substituted L, and for A, L^, which are the 

dimensions of length and of area respectively, giving 

1^ Xp=LT-\ or p=VT-\ 

which are the dimensions of specific resistance. 



RESISTANCE 69 

In tables of specific resistance it is usually expressed in 
microhms for solids and in ohms for solutions Often it is 
expressed in C.G.S. units of the electro-magnetic or electrostatic 
system. 

Example. Assume the specific resistance of tin to be 13.36 
microhms. What is the resistance of a tin wire 350 feet long 
and i inch diameter ? 

Solution. The cross section of the wire is 3.1416 X (1/16)^ 
= 0.0123 square inches. 0.0123 X 6.45 = 0.0793 square centi- 
meter. The length of the wire is 350 X 30.48 = 10,668 centi- 
meters. Substituting in expression (i) the values as above gives 

T. . . 10,668 „ . , 

Resistance = 13.36 X ~ = 1,797,282 microhms = 1.797 

0.0793 

ohms. 

If the diameter, d^ of a wire is given instead of the cross- 
sectional area, the expression for the area of the cross section 
will be 7r(^V4> because (/^/4 is the square of (i/2. As the resist- 
ance of a conductor varies inversely with its cross-sectional area, 
it will vary with ^/7:d\ which reduces to 1.2732/^^. Multiplying 
this by the length and by the specific resistance gives the 
expression 

sp. res. X 1.2732 X / . ) 

d' ^ ^ 

for the resistance of a wire of length / and of diameter d. 

Example. What is the resistance of a copper wire i^ meters 
long, I millimeter thick, the specific resistance of copper being 
taken as 1.652 microhms ? 

Solution. Substituting in the formula for / and d their 

values in centimeters and for sp. res. its value 1.652 we have 

_, . ^ 1.652 X 1.2732 X ISO . I 

Resistance = — ~ ~ = 3i)55o microhms, or 

o.oi 

0.03,155 ohm. 

In calculating the resistance between electrodes in a solution 



70 ELEMENTARY ELECTRICAL CALCULATIONS 

the facing areas are the only ones usually taken into account, 
because the rear or remote faces are of comparatively little 
value in the conduction. The conductor is treated as a prism 
or, where the electrodes are of unequal size, as a frustum of a 
pyramid the area of whose bases is determined by the faces of 
the electrodes. 

Example. An electroplater's bath has a specific resistance 
of 42 ohms. The electrodes are each 140 square inches in 
area and are 3 inches apart. Calculate the resistance. 

Solution. The area of the electrodes is 140 X 6.45 = 
903 square centimeters. The distance separating them is 
3 X 2.54 = 7.62 centimeters. Substituting these values in 
expression (i) we have 

7 62 
i? = 42 X -^ — = o.s'54 ohm. 
903 

Example. In a battery the facing areas of the plates are 
12 and 40 square inches respectively and they are f inch 
apart. The specific resistance of the solution is 9 ohms. Cal- 
culate the resistance. 

Solution. The average area of the plates is = 

2 

26 square inches = 167.7 square centimeters, f inch = 1.905 

centimeters. Substituting in the formula as before gives 

R = qX ', ■ = 0.102 ohm. 
167.7 

Calculations of the resistance of solutions are not of much 
accuracy. They may be useful as approximations. The least 
change in the composition of a solution affects its specific resist- 
ance. The approximate formulas given above sufl&ce for 
most cases. 

Circular Mils. A mil is a measure of length; it is one one- 
thousandth of an inch. 



RESISTANCE 7 1 

A circular mil is a measure of area ; it is the area of a circle 
one mil in diameter. 

The area of a circle may be expressed in circular mils. The 
area of a circle in circular mils is equal to the number of circular 
mils expressed or obtained by squaring its diameter given in 
linear mils. Thus the area of a circle 3 mils in diameter is 
equal to (3)^ circular mils, or 9 circular mils. 

The diagram shows a circle supposed to be of 3 mils diameter. 
By geometry we know that the area of the circle is 0.7584 
that of the circumscribing square. Assume the square to be 
divided into 9 small squares, each one supposed to be i mil 
in diameter, and whose united areas, 
or the sum of whose areas, is equal 
to the area of the large square. These 
squares are equal in number to the square 
of the diameter of the circle in circular 
mils. Within each small square is in- 
scribed a circle whose diameter, under 
the conditions of the proposition, is 
I mil. The ratio of the area of each small circle to its circum- 
scribing square being the same as that of the large circle to 
its circumscribing square, and the sum of the areas of the small 
squares being equal to the area of the large square, it follows 
that the area of the large circle is equal to the sum of the areas 
of the small circles. But the number of these is equal to the 
square of the diameter of the large circle in mils. The area 
of each small circle is equal to i circular mil. Hence it follows 
that the area of the large circle is equal to a number of circular 
mils, equal to the diameter in mils squared. What is true for 
a circle of one size is true for any other, so that the proposi- 
tion is proved. 

The cross-sectional area of wires and circular conductors is 
often expressed in circular mils. In many cases it is the 




^2 ELEMENTARY ELECTRICAL CALCULATIONS 

most convenient way of working and is in constant use by 
engineers. 

A copper wire i mil in diameter and i foot long has a resist- 
ance at the temperature of 75° F. (24° C.) of 10.79 ohms. This 
figure is necessarily approximate, because the resistance of 
different samples of copper varies, but it is in constant use in 
practical work. 

What has been explained gives us the rule: 

The resistance of a copper wire is equal to the product of its 
length in feet by 10.79 divided by the square of its diameter in 
mils. 

^ ^ / X 10.79 
d^ (circ. mils) 

The square of the diameter in mils of a circle is its actual area 
in circular mils. 

Example, What is the area of a circle } inch in diameter, 
in circular mils? 

Solution. I inch = 1,000 X i mils =750 mils. This is 
the diameter of the circle in linear mils. Its area in circular 
mils is equal to (750)^ = 562,500 circular mils. 

Example. Calculate the resistance of a copper wire 1,034 feet 
long and 320 circular mils in area. In such statements as this, 
by " area" cross-sectional area is always understood. 

Solution. Substituting the given values in the formula we have 

R^ 1,034X10.79 ^ 86 hms. 
320 

This can be done by logarithms. 

log. 1,034 3-01452 

log. 10.79 1-03302 

colog. 320 7.49485—10 

log. 34.865 154239 



RESISTANCE 73 

Example. Calculate the resistance of a copper wire 14 mils 
diameter and 1,644 feet long. 

Solution. (14)2 = 196, the area of the wire in circular 

mils. Substituting as before gives 

„ 1,644 X 10.79 u 

^ ^ ' — U' = 90. <; ohms. 

196 ^ ^ 

By logarithms : 

log. 1,644 3-21590 

log. 10.79 1-03302 

colog. 196 7.70774—10 



log. 90.5 1.95666 

Efifect of Temperature of Conductors on their Resist- 
ance. The resistance of conductors varies with the tem- 
perature, increasing as the temperature rises. Tables of the 
increase are published; the rate of increase varies with the 
temperature, the rate decreasing as the conductor's tempera- 
ture rises. If the resistance of a copper conductor at 0° C. be 
taken as i, resistance at 20° C. will be 1.07968; at 50° C, 
1.20625; at 80° C, 1.33681. These figures give as percent- 
age rates for 1° change in temperature between 0° C. and 
20° C. 0.3984; between 20° C. and 50° C. 0.3856; between 
50° C. and 80° C. 0.3607 if the resistances at 0°, 20°, and 50° 
be taken as 100% for each case. These figures are the results 
of experiments on a certain sample of copper. Other samples 
would give different results. 0.40 per cent is a convenient 
approximate coefficient. 

Example. If the resistance of a copper wire at 10° C. is 29 
ohms, what will its resistance at 19° C. be ? 

Solution. The change in temperature is 19° — 10° = 9°. 
At the temperature of the problem the percentage may be 
taken as 0.3984, and multiplying this by the number of degrees 



74 ELEMENTARY ELECTRICAL CALCULATIONS 

gives the total change as a percentage, or 9 X 0.3984 = 3.586. 
This percentage has to be added to the original resistance, 
29 ohms. Thus 29 X 1.03586 = 30.04 ohms. 

Example. What heat is indicated by a copper wire whose 
resistance at 15° C. is 300 ohms when its resistance rises to 495 
ohms as it is placed in a chimney ? Take 0.38 as the percent- 
age of change. 

Solution. It rises -^ = 65 per cent in resistance. Dividing 

the per cent increase in resistance by the per cent increase for 
i°C. gives the number of degrees C. it has risen. Thus 65 -^■ 
0.38 = 171° C, to be added to the original temperature; 15° 
+ 171° = 186° C. 

Example. A mile of copper wire in the open air varies 
from 7.9 ohms at night to 8.6 ohms in the daytime; what is the 
difference in temperature ? 

Solution. The increase in resistance is -^ = 8.86 per cent. 

Taking 0.39 as the per cent per degree, and dividing as before, 
we have 8.86 -^ 0.39 = 22.7° C, the increase in temperature. 

The following is Matthiessen's formula : 
Resistance at temp. t° C. = resistance at 0° C. X (i + a^ + hf). 
The following are values of a and h : 

a b 

Pure copper, +0.003,84 +0.000,001,26 

Mercury, 0.000,748,5 —0.000,000,398 

German silver, 0.000,443,3 +0.000,000,152 

PROBLEMS. 

Calculate the resistance of two conductors in parallel, one of 
5 ohms and the other of 9 ohms resistance. Ans. ^-^^ ohms. 

What is the combined resistance of two conductors of ^ and ^ 
ohm in parallel ? Ans. ^^ ohm. 

Combine the resistances J ohm and -J ohm in parallel. 

Ans. 1^ ohm. 



. RESISTANCE 75 

Calculate the resistance of f and 9 ohms in parallel. 

Ans. -^^ ohm. 
Combine the following resistances in parallel, f , ^, and -^-^ ohm. 

Ans. ^f I ohm. 
Calculate the resistance of 4 conductors in parallel of i, 3, 7, and 
12 ohms resistance respectively. Ans. -^^ ohm. 

What is the resistance of three wires in parallel of conductances 
of 5, 16, and f ohms respectively? Ans. j^-^ ohm. 

A resistance coil has a resistance of 1,015 ohms. A resistance is 
to be placed in parallel with it so as to reduce the combined resist- 
ance to 1,000 ohms. What must the second resistance be? 

Ans. It is the reciprocal of the difference of the reciprocals, 
67,666 ohms. 

What is the resistance of a wire of 4 ohm in series with three 
wires in parallel of i, ^\, and J ohm respectively, and with 
another conductor in series of 16 ohms resistance? 

Ans. i6^^\ ohms. 
The resistance of 1,000 feet of wire o.oi inch in diameter is 105.641 
ohms. What is the resistance of 1,000 feet of wire 0.008 inch in 
diameter? Ans. 1.65 ohms. 

What is the resistance of 633 feet of wire 0.08 inch in diameter if 
the resistance of 1,000 feet of wire 0.018 inch in diameter is 33.135 
ohms? Ans. 1.061 ohms. 

1,000 feet of No. 15 wire weigh 9.84 pounds, 1,000 feet of No. 10 
wire weigh 31.38 pounds, with a resistance of 1.023 ohms. What is 
the resistance of the 1,000 feet of No. 15 wire ? Ans. 3.26 ohms. 

Compare the resistance of two conductors of 6,000 and 9,000 cir- 
cular mils area. Ans. i to |. 
Assume three conductors of 5, 6, and 8 ohms resistance respec- 
tively. A current of i ampere flows through them. Calculate its 
distribution, Ans. -^-^-^ ampere in 5 ohm lead. 

y\Og- ampere in 6 ohm lead. 
y3y0^ ampere in 8 ohm lead. 
Calculate the distribution of one ampere of current in 4 con- 
ductors in parallel of J, i, 7, and 9 ohms resistance respectively. 

Ans. i|^ ampere in J ohm lead, 
fif ampere in J ohm lead. 
3^1^ ampere in 7 ohm lead. 
■g^jj ampere in 9 ohm lead. 



76 ELEMENTARY ELECTRICAL CALCULATIONS 

1,500 feet of conductor of 6,530 circular mils section have a resist- 
ance of 2.439 ohms. What is the resistance of 1,700 feet of 5,178 
circular mils conductor? Ans. 3.48 ohms. 

Two wires of identical cross section are 19,173 and 29,871 feet 
long respectively. The longer wire is of 13. i ohms resistance. 
Calculate the resistance of the other. Ans. 8.41 ohms. 

Compare the resistances of equal lengths of wire of 1,021 and 
8,234 circular mils. 
Ans. The small one has 8.06 times the resistance of the large one. 

Calculate the resistance of a bar of copper, sp. r. 1.652; 15 feet 
long and 2 square inches cross section. 

Ans. 58.53 microhms. 

5,853 X 10-8 ohms. 

Calculate the resistance of a copper wire 114 feet long and o.oio 
inch in diameter, taking sp. r. as 1.652. 

Ans. 11,332,000 microhms. 
11.332 ohms. 

A copper wire has a resistance at 16° C. of 257 ohms. Calculate 
the resistance at 27° C, taking the coefficient as 0.40 per cent. 

Ans. 268.3 ohms. 

The resistance of a copper conductor changes from 30 ohms to 
27.9 ohms at an ordinary temperature. What is the fall in tempera- 
ture? Ans. 17.5° C. 

If the resistance of a copper conductor at 50° F. is 295 ohms, what 
is its resistance at 60° F. ? (50° F. = 10° C. ; 60° F. = 15.5° C.) 

Ans. 301 ohms. 

Calculate the relative resistance of German silver at 11° C. and 
at 15° C, and determine the change in resistance per degree C. 
Take the resistance at 0° C. as unity. 

Ans. At 11° C. 1.004,89; at 15° C. 1.006,68. Increase for 4 degrees 
0.001,79; fo^ I degree 0.000,45. 

What is the resistance at 19° C, of a copper wire whose resistance 
at o°C. is 17 ohms? 

Applying Matthiessen's formula we have 

Resistance at 19^ C. = 17 ohms X [i -f (0.003,84 x 19) 

+ (0.000,001,26 X 19^)] = 18.24 ohms. 



CHAPTER VI. 

KIRCHHOFF'S LAWS. 

Kirchhoff's Laws. Kirchhoff*s laws apply to the distribu- 
tion of current and e.m.f. in branching conductors and network 
of conductors when they are passing steady currents. The laws 
are two in number. 

1. When conductors forming parts of an active circuit meet 
in one point the algebraic sum of the currents is zero, negative 
signs being assigned to currents leaving the junction and posi- 
tive signs to those going to the junction, or vice versa. Or put 
into simpler language, the currents going to a junction are 
equal in the sum of their intensities to the sum of the intensities 
of the currents leaving it. 

2. When conductors form a circuit comprising within it a 
source of e.m.f., the sum of the products of the intensity of the 
current within each part of the circuit into the resistance of 
the same part is equal to the e.m.f. of the system provided 
the elements of current be taken in cyclical order. 

In other words, Kirchhoff's second law applies when the direc- 
tion of the currents can be deduced. Otherwise, although it is 
true, it cannot be used in practice. 

Kirchhoff's laws in calculations are used conjointly with 
Ohm's law. Some simple examples of their use are given here. 

Example. Three conductors meet in a point. One con- 
ductor carries a current of 3 amperes to the junction, another 
carries a current of 6 amperes away from the junction. 
What current does the third conductor carry and in what 
direction ? 

Solution. This case comes under the first law. As the cur- 

77 



78 ELEMENTARY ELECTRICAL CALCULATIONS 

rents going to the junction must equal those leaving it, the third 
conductor carries a current of 3 amperes to the junction; thus 
the currents to the junction are 3 + 3=6 amperes and the 
current from the junction is thus equal to them. 

To do it by algebraic addition, call the current to the junction 
+ 3. Call the current from the junction — 6. Let x be the 
other current. Then by the law we have 

rx: + 3 — 6 = o and x = ^. 

As the sign of x is positive the current is to the junction. 

Example. Calculate the following 
^ -^^^^ circuit by Kirchhoff's laws. In the dia- 
gram the capital letters indicate the 
currents in each branch. The diagram 
may be taken as explaining the circuit 
and dispensing with the necessity of an 
explanation. 

Solution. Let 5, ^, q indicate the 
resistances of the respective branches, h including the resistance 
of the battery. 
From the first law is obtained the expression 

P + Q = 5. (i) 

The e.m.f. of a circuit is equal to the product of the current 
by the resistance. (Ohm's law.) In the circuit under consider- 
ation the portions B and Q constitute one cycle and the por- 
tions B and P another, as it is evident that a consecutive circuit is 
afforded by each of the two divisions specified. This is what is 
meant by " taking the elements of current in cycUcal order." 
Applying the second law we obtain 

Bb + Pp = e.m.f. of the generator. (2) 

Bb + Qq = e.m.f. of the generator. (3) 



y 



KIRCHHOFF'S LAWS 



79 



In general the sum of the R.I. drops of either of the cycles is 
equal to the e.m.f. of the system. 

This follows from Ohm's and Kirchhoff's laws, as each cycle 
includes the battery and a full lead from terminal to terminal. 
The potential difference between the ends of the lead P is the 
same as that between the ends of the parallel lead Q. 

Let the resistances be as follows: 6 = 23;^ = 7;^ = ii. 
Let the e.m.f. be i. Substituting these values in equations (2) 
and (3) and writing them out as simultaneous equations with 
equation (i) as one of the group we have 

P + Q==B. (4) 

23^+7^ = 1. (5) 

23^+11(3 = 1. (6) 

Solving these by the ordinary method for simultaneous 
equations we find 

P = 0.0224; Q = 0.0142; and B = 0.0366. 

To do it by determinants write the equations as follows : 

27,B + oP + II (2 = I. 

235+7P+ oQ = i. 
-B+ P+ (2=o. 

Arranging the determinants for the value of B gives 

I o II 23 o 

17 o -- 23 7 
01 I — I I 

The determinant dividends for P and Q respectively are 

o 

7 



II 
o 

I 



— = 0.0366 ampere = B. 



49 



23 
23 
- I 



II 
o 

I 



II and 



23 
23 
-I 



Giving 



II 



P = — =0.0224 ampere, and Q 
491 



_7_ 
491 



0.01426 ampere. 



So ELEMENTARY ELECTRICAL CALCULATIONS 



Example. Calculate the circuit of the cuts by Kirchhoff's 
laws. The capital letters indicate currents, the figures indi- 
cate the resistances. Let e.m.f. = i. 

Solution. The first law gives the following equations : 

-^ B- R- S =o. (i) 

S - P -Q =o. (2) 

T-P-Q=o. (3) 

From the second law are found the 
following : 




45 



6B + 3R = i. 
2 e - s r = o. 
2 (2 - P =0. 



Subtracting (2) from (3), 
T 



(4) 
(S) 
(6) 

(7) 



The equations (i), (2), (4), (5), (6), and (7) are solved as 
simultaneous equations, giving 



6 

315 


R- ^, 

315 


315 


315 


315 
PROBLEMS. 


B=^. 
315 



Three conductors meet in a point or junction. One carries a current 
of 21 amperes to the junction, another carries a current of 5 amperes 
to it. What is the current in the other conductor? 

Ans. A current of 26 amperes flowing from the junction. 

Five conductors meet in a point. Three carry currents of 6, 8, and 
9 amperes respectively to the junction; one of the other conductors 
carries a current of 39 amperes away from the junction. What is the 
current in the remaining conductor? 

Ans. A current of 16 amperes to the junction. 



KIRCHHOFF'S LAWS 8 1 

Three conductors a, b, and c meet in a point; a carries a current of 
7 amperes away from the junction, b carries a current of 9 amperes 
and c one of 2 amperes. What is the direction of the currents in b and 
c? Ans. Current in b to the junction; current in c away from it. 

In the last case the current in b rises to 11 amperes and in c falls to 
I ampere. What is the current in a? 

Ans. 10 amperes from the junction. 

In the same case c rises to 5 amperes, a remaining unchanged. What 
happens to 6? Ans. It becomes 15 amperes to the junction. 



CHAPTER VII. 



ARRANGEMENT OF BATTERIES. 

Electro-motive Force of a Battery. — Resistance of a Battery. — Potential 
Drop of a Battery. — Greatest Current from a Battery. — Rules for 
Calculating a Battery. — Energy Expended in a Battery. — Rule 
for Calculating a Battery of Given Efficiency. — Discussion. — 
Problems. 

Electro-motive Force of a Battery. The electro-motive 

force of a battery is equal to the sum of the electro-motive forces 
of its cells in series, irrespective of the number in parallel. Thus 




a battery arranged lo in series and 5 or 6 or any other number 
in parallel has the e.m.f. of the sum of the e.m.f.'s of the 10 cells. 
The first of these cuts represents a battery of 3 cells connected 
in series. The next represents a battery also of 3 cells con- 
nected in parallel. The third represents a battery of 9 cells, 
arranged 3 in series and 3 in parallel. 

Resistance of a Battery. The resistance of a battery 
arranged like a rectangle is equal to the sum of the resistances 
of the cells in series divided by the number of cells in parallel. 

Example. Suppose a battery to consist of 50 cells arranged 

82 



ARRANGEMENT OF BATTERIES 83 

5 in series and 10 in parallel. What are its e.m.f. and resistance? 
Each cell has an e.m.f. of i.i volt and a resistance of 4.2 ohms. 

Solution. The e.m.f. of the battery is equal to the e.m.f. of 
a single cell multiplied by the number in series; i.i X 5 = 5.5 
volts. The resistance of the battery is equal to the resistance 
of a single cell multiplied by the number in series and divided 
by the number in parallel; 4.2 X 5 -^ 10 = 2.1 ohms. 

Potential Drop of a Battery. The potential drop some- 
times called the RI drop of a battery, or lost volts, is the 
difference between the e.m.f. on open circuit and the potential 
drop between its terminals when connected by a conductor of 
given resistance. The RI drop of the battery varies according 
to the resistance of the outer circuit or conductor connecting its 
terminals. It is the e.m.f. expended on maintaining the current 
through the resistance of the battery. 

Example. A voltmeter connecting the terminals of a battery 
on open circuit reads 24 volts. When the terminals are connected 
by a conductor the voltmeter shows 20 volts. What is the RI 
drop of the battery when its terminals are connected by the 
conductor in question ? 

Solution. It is 24 — 20 = 4 volts. 

The drop of potential of a battery is equal to its e.m.f. mul- 
tiplied by its resistance and divided by the resistance of the 
entire circuit. 

Example. A battery of 3 volts and 5 ohms is connected in cir- 
cuit with a resistance of 21 ohms. What is its drop of potential? 

Solution. By the rule just given it is equal to 3 X 5 -^ 
(5 + 21) = 0.577 volt. 

Greatest Current from a Battery. If a given number of 
cells or equivalent generators are to be used to produce a current 
through a given resistance, they will produce the greatest cur- 
rent when the external and internal resistances are the same. 

Assume 40 cells each of i ohm resistance and i volt e.m.f. 



84 ELEMENTARY ELECTRICAL CALCULATIONS 

and assume an external resistance of 5 ohms. If the cells are 
in series their e,m.f. will be 40 volts and their resistance will 
be 40 X 1/2 = 20 ohms. The current in the circuit will be 

1. 6 amperes. 



20+5 
If the cells are 20 in series and 2 in parallel, their e.m.f. will 

be 20 volts and their resistance 5 ohms, which is that of the 

20 

outer circuit. The current will be = 2 amperes. 

5 + 5 
If we go a step further and assume the cells to be 10 in series 

and 4 in parallel, the internal resistance is 1.25 ohms and the 

e.m.f. is 10 volts. The current will be =1.6 amperes. 

1.25 + 5 
These three cases illustrate the law enunciated above. The 

greatest current is produced when the battery is so connected 

as to have its internal resistance equal to that of the outer circuit. 

Rules for Calculating a Battery. A general method for 
calculating the number and arrangement of cells of battery to 
maintain the greatest current through a given resistance can be 
based upon the above principle. 

Arrange the cells so as to give twice the e.m.f. required and 
so as to have a resistance equal to that of the external circuit. 
To effect this it will often be necessary to arrange the cells irregu- 
larly. The more regularly the cells are arranged the better will 
the results be, as a rule. 

Example. The constants of a cell of a battery are i volt 
and 4 ohms. The battery is to maintain a current of 0.56 
ampere through a circuit of 30 ohms resistance. Calculate the 
number and arrangement of the cells. 

Solution. The e.m.f. required for the outer circuit will be, 
by Ohm's law, 0.56 X 30 = 17 volts. As the internal and 
external resistance are to be equal to each other, the battery, 
by the law of the distribution of energy, must maintain double 
this e.m.f., 17 volts to be expended within itself and 17 volts 



ARRANGEMENT OF BATTERIES 85 

to be expended upon the outer circuit. Therefore 34 cells in 
series will be needed. A single series of 34 cells would have a 
resistance of 34 X 4 = 136 ohms. Four series in parallel 
would have one-fourth this resistance, or 34 ohms. This would 
be a close enough approximation for ordinary purposes. Five 
series in parallel would have a resistance of 136 -r- 5 = 27.2 
ohms, again a close enough approximation for most purposes. 
If a group of 20 in series and 5 in parallel is placed in series 
with a group of 14 in series and 4 in parallel, the resistance of 
the battery will be the sum of the resistances of the two groups. 
The resistance of the first group will be (20 X 4) -j- 5 = 16 ohms. 
That of the second group will be (14 X 4) -^ 4 = 14 ohms. 
The total resistance of the battery will be 16 + 14 = 30 
ohms. The number of cells required is (5 X 20) + (4 X 14) =156. 

A good general rule is the following : 

To calculate the number of cells to produce a given current 
through a given resistance, group enough cells in parallel to 
give on short circuit twice the required current. Treating 
this as a single cell place enough of these groups in series to 
give twice the potential drop of the outer circuit. 

Example. A current of 5 amperes is to be maintained through 
a resistance of 6 ohms. The battery constants are i volt and 
4 ohms. Calculate the cells. 

Solution. Twice the required current is 10 amperes. As 
the resistance of a single cell is 4 ohms, 40 cells in parallel will 
have a resistance of yV ohm. The voltage of a single cell is i ; 
the group of 40 cells in parallel will give on short circuit a 

current of i -^ — =10 amperes. The drop of the outer cir- 
10 

cuit is 5 X 6 = 30 volts. Twice this is 60, so that 60 groups 
are needed in series to give the e.m.f. The total number of 
cells is 60 X 40 = 2,400. 

Testing the correctness by Ohm's law, we have the external 



86 ELEMENTARY ELECTRICAL CALCULATIONS 

resistance 6 ohms, the internal resistance 6 ohms, the e.m.f. 

6o volts, and current = = 5 amperes, showing the cor- 

+ 

rectness of the work. 

Energy Expended in a Battery. The energy expended in 
any part of a circuit is porportional to its resistance. The 
energy expended in a battery forming part of a circuit and main- 
taining a current in it is proportional to its resistance also. 
This energy is wasted. 

The proportion of energy wasted in a battery supplying a 
circuit is equal to the resistance of the battery divided by the 
total resistance of the circuit. 

Example. A battery of 13 ohms resistance supplies a cir- 
cuit of 19 ohms external resistance. Calculate the proportion 
of energy wasted. 

Solution. The total resistance of the circuit is 13 + 19 = 
32 ohms. This is the divisor of the fraction whose numerator 

is 13. The energy wasted in the battery is -^ =40.6 per cent. 

Rule for Calculating Battery of given Eflaciency. To 

calculate the battery to supply an external circuit of given 
resistance with a given current at a given percentage of loss, 
proceed as follows. As the drop in potential is equal to the 
product of the resistance of the part of the circuit in question 
multiplied by the current, the energy expended in a battery 
is proportional to its drop in potential, the same as it is pro- 
portional to its relative resistance. Subtract the per cent of 
loss of energy in the battery from 100. Take the difference as 
the denominator of a fraction whose numerator is the per cent 
of loss. The resistance of the external circuit multiphed by 
this fraction is the resistance of the battery. The potential 
drop of the external circuit multiplied by the same fraction 
is the potential drop of the battery. The e.m.f. of the battery 



ARRANGEMENT OF BATTERIES 87 

is the sum of the potential drops. Or multiply the total resist- 
ance of the circuit by the current. The product will be the 
e.m.f. of the battery. 

Example. Calculate the resistance and e.m.f. of a battery 
to maintain a current of 3 amperes through a resistance of 18 
ohms with an expenditure of 19 per cent of energy in the battery. 

Solution. The resistance of the battery is given by the 
product of the resistance of the outer circuit by the fraction 

^ — =12 , iS X — =4.22 ohms. This is the resistance 

100 — 19 81 81 

of the battery. The total resistance of the circuit is 18 + 4.22 

= 22.22. The e.m.f. of the battery is equal to the product of 

the total resistance by the current, 3 amperes, which is 22.22 

X 3 = 66.66 volts. The last result can be reached by the 

other method. The potential drop of the outer circuit is equal 

to the product of the current by the resistance. It is therefore 

18 X 3 = 54 volts. 54 X r^ = 12.66, the potential drop of 

81 

the batter}^ The e.m.f. of the battery is 12.66 + 54 = 66.66 

volts, as before. 

The following is a somewhat shorter rule. Subtract the 
per cent of loss from 100 per cent, which gives the per cent of 
efficiency. Divide the potential drop of the external circuit by 
the efficiency expressed as a decimal; the quotient is the e.m.f. of 
the battery. Subtract the potential drop of the external circuit 
from that of the e.m.f. of the battery; the remainder is the 
potential drop of the battery. 

Example. A current of 29 amperes is to be maintained 
through II ohms resistance with 29 per cent loss. Calculate 
the e.m.f. of the battery. 

Solution. 100 — 29 = 71, the efficiency. 29 X n =319? the 
potential drop of the outer circuit. 319 -f- 0.71 = 450, the e.m.f. 
of the battery. 450 — 3 19 = 131, the potential drop of the battery. 



S8 ELEMENTARY ELECTRICAL CALCULATIONS 

Discussion of Principles of Calculating Batteries. The 

whole matter of calculating a battery to give the exact resistance 
and electro-motive force demanded by any given conditions is 
rather theoretical than practical. The resistance and electro- 
motive force of a battery continually change ; the resistance may 
increase or diminish, the e.m.f . generally decreases as the battery 
is in active service. The oxidation of the hydrogen of the water 
molecule in a battery is termed depolarizing. This is an essen- 
tial part of the action, and is often interfered with when a current 
is taken from a battery. In such case a battery is said to be 
polarized. Even standing on open circuit is liable to change a 
battery's constants. It follows that a calculation correct for a bat- 
tery in good condition rapidly becomes incorrect, as the battery 
changes its constants as a current is taken from it or as it stands 
on open circuit. The following principles are adapted to calcu- 
lating the cells of a battery of given constants required to main- 
tain a given current through a given resistance. Let 

n = number of cells in series. 

r = resistance of a single cell. 

e = e.m.f. of a single cell. 

R = resistance of outer circuit. 

/ = current required. 
The total e.m.f. of the circuit will be that of a single cell 
multiplied by the number of cells in series, which is equal to ne. 
The resistance of the battery is the resistance of a single cell 
multiplied by the number of cells if the cells are in series, or is 
nr. The total resistance of the circuit is the sum of the resist- 
ance of the battery and of that of the outer circuit, or is nr + R. 
By Ohm's law the current is equal to the e.m.f. divided by 

the resistance, or r ^ ^^ /j\ 

nr+ R' ^ ^ 

Multiplying both members by nr -\- R gives 

nrl -{- RI = ne^ (2) 



ARRANGEMENT OF BATTERIES 89 

and transposing, ^j = ^^ - nrl. (3) 

Dividing throughout by e — ;'/ and transposing, 

n = -' (4) 

e — rl 

For this formula to be applicable to single cells in series it is 
evident that rl must be less in value than e. Otherwise the 
denominator will reduce to zero, giving infinity as the value 
of n, or in words stating that an infinite number of cells 
would be required unless the e.m.f. of a single cell exceeds the 
potential drop of a cell at the given current. 

To meet this condition group two or more cells in parallel 
and treat each group as a single cell. The e.m.f. of the group 
is the same as that of a single cell; the resistance is that of a 
single cell divided by the number in the group. If m cells are 
grouped in parallel, we have e (r/m X /) for the denomi- 
nator of the expression. If the value of this expression is 
twice the value of the required current, enough cells are in 
parallel to make it possible to obtain the current. 

From the above considerations is found the smallest number 
of cells in parallel which will give a stated current. The mini- 
mum or smallest number of cells required is given if the internal 
resistance is equal to the external, carrying with it the condition 
that the e.m.f. of the battery, or ne, shall be twice the drop of 
the outer circuit. This drop is RI, so the condition is ex- 
pressed in the equation 

ne = 2 RI. (5) 

Multiplying the second term of (4) by e and equating it with 
the second member of (5) gives 

.RI=-^. (6) 

e — rl 

Multiplying by e — Ir and dividing by 2 RI gives 

e^rl=-j (7) 



90 ELEMENTARY ELECTRICAL CALCULATIONS 
and transposing and reducing we find 

rI=e-^ = '- (8) 

2 2 

and - = 2 7, (9) 

r 

which last equation expresses the condition for the smallest 
number of cells for a given current. The rule is thus expressed 
in words : 

To obtain the smallest number of cells for a given current 
group enough cells in parallel to give on short circuit twice the 
required current. Treating the group as a single cell apply 
formula (4), when n will be the number of groups to be put in 
series. The total number of cells is the product of the num- 
ber in a group by the number of groups in series. 

Example. A current of 5 amperes is to be taken from a 
battery of cell constants 2 volts and ^^ ohm. The external 
resistance is 2 ohms. Calculate the number of cells required. 

Solution. Applying (9) we have — = 2 ^ -^ = 6.66. 

r 10 

As this is less than twice the current, the series arrangement 
will be disadvantageous, though possible, as requiring the 
greatest number of cells all in series and therefore giving the 
highest resistance. If two cells are placed in parallel the resist- 

ance of the group will be ^^0 ohm, and — = 2 -4- ""^=13.33, 

which is more than twice the current required. Treating this 
group as if it were a single cell and applying (4) we find for the 
groups in series 

2 - (3/20 X 5) 25/20 

The total number of cells is the number in parallel multiplied 
by the number in series, or 2 X 8 = 16. 
Testing by Ohm's law, the resistance of the battery being 



ARRANGEMENT OF BATTERIES 9 1 

-^ X 8 -^ 2 = — the resistance of the outer circuit being 2, 
10 20 

and the e.m.f. of the battery being 2 X 8 = 16, the current 

is given by the formula, current = — ; = 5, which is the 

24/20 + 2 

current required, thus proving the correctness of the operation. 

' Example. If possible arrange cells of the above constants 
in a series so as to give the same current through the same 
resistance. 

Solution. Substituting the values of the problem in the 

denominator of (4) gives e — rl = 2 — f-^ X 5 ) = 0.5. It 

is therefore possible to carry out the series connection. The 
number of cells, is by (4), 

5 X2 _io_ 

— _ = 20. 

2 - (3/10 X 5) 0-5 
Testing as before by Ohm's law we have 

20 X 2 40 

current = , ; — = -— • = <. 

(20 X 3/10) + 28 ^* 

proving the correctness of the operation. 

The last arrangement is uneconomical in every respect. The 
two examples illustrate the advantage of proper connection of 
cells. 

Example. Assume the figures of the last two problems, 
except that the resistance of the battery is f ohm per cell. 
How many cells in series would give the current ? 

Solution. The denominator of the second term of (4) 

becomes 2— (— X5)=o. Therefore an infinite number of 

cells would be needed. The current cannot be supplied by 
any number whatever of the cells in series. 



92 ELEMENTARY ELECTRICAL CALCULATIONS 

PROBLEMS. 

Calculate the e.m.f. and resistance of a battery of 486 cells 27 in 
series and 18 in parallel, each cell having 0.05 ohm resistance and 
1.97 volt e.m.f. Ans. 0.075 ^^^ ^^^ 53-^9 volts. 

The constants of a battery cell are 1.8 volts and y ohm. Arrange 
such cells for 3 amperes current through 21 ohms external resistance. 

Ans. 70 cells in series and 2 in parallel. 

Calculate a battery of cell constants i volt and 4 ohms to give a 
current of 5 amperes through a resistance of 5 ohms. 

Ans. 50 cells in series and 40 in parallel. 

Calculate the number of cells of 1.7 volt and 0.3 ohm each to main- 
tain a current of 2.3 amperes through a resistance of 21 ohms with an 
efficiency of 75 per cent. 

Ans. One group 2 in parallel and 28 in series, and another group 
10 in series, the two in series with each other. 

125 cells of 1.75 volts and f ohm each are arranged 5 in parallel and 
25 in series. There is an external resistance of 3.4 ohm. Calculate 
the battery constants and the current. 

Ans. 43.75 volts; 3.75 ohms; 6.12 amperes. 

Calculate the resistance of a battery to supply four magnets in 
parallel, the coil of each magnet being of 4.6 ohms resistance and the 
efficiency to be 90 per cent. Ans. 0.125 ohm. 

Arrange cells of i.i volts and 6 ohms to supply above magnets with 
2j-ampere current. Ans. 2 cells in parallel will give 2.27 amperes. 

What is the resistance and e.m.f. of a battery 11 in parallel and 7 
in series with cell constants 1.5 volts and 0.25 ohm? 

Ans. 0.159 ohm; 10.5 volts. 

Calculate the same factors if the above battery is in parallel. 

Ans. 0.00325 ohm; 1.5 volts. 

Calculate the same for the battery in series. 

Ans. 19.25 ohms; 115. 5 volts. 

What current will each of the arrangements give through 11 ohms 
external resistance ? Ans. First arrangement, 0.94 ampere. 

Second arrangement, 0.136 ampere. 
Third arrangement, 3.8 amperes. 



ARRANGEMENT OF BATTERIES 93 

A current of 3 amperes is to be produced through a wire of 30 ohms 
resistance, with a battery of 2 volts and 0.2 ohm cell constants. Cal- 
culate the cells for 80 per cent efficiency. 

Ans. Approximately 59 in series and 2 in parallel. 

What is the exact efficiency of the above arrangement? 

Ans. 83.3 per cent. 

What current will be given by 51 cells, each cell of 1.07 volts 3 
ohms, connected in series, through an external resistance of 19 ohms ? 

Ans. 0.317 ampere. 

If 30 cells of the above battery are arranged 3 in parallel and 10 in 
series, what current will they give through the same resistance ? 

Ans. 0.369 ampere. 

What will a single cell of the same battery give through the same 
resistance? Ans. 0,486 ampere. 

Note. — This is a case where a single cell gives more current than a 
number of cells. 



CHAPTER VIII. 

ELECTRIC ENERGY AND POWER. 

Potential. — Proof of Numerical Value of Potential. — Potential Drop or 
(see page 95) R. I. Drop. — Electric Energy. — Practical Unit of 
Electric Energy. — Electric Power or Activity. — Practical Unit of 
Electric Power. — Relations of Power to Current, Resistance, and 
e.m.f . — Equivalents of the Watt-Problems. 

Potential. Potential is of various kinds, affecting mass, 
electric quantity, and other things, each kind of potential affect- 
ing only one of them. It can be defined as a condition of a 
point in space such that the energy involved in moving a unit 
mass or quantity from the point to an infinite distance or from 
an infinite distance to the point would be numerically equal to 
the potential. 

The energy involved in the moving of a unit mass or the unit 
quantity from a point of one potential to that of another is 
numerically equal to the difference of the two potentials. 

Electric potential affects electric quantity only. 

Proof of Numerical Value of Potential. To prove the 
above, let e represent a charge of electricity at a point in space. 
Let a unit quantity be acted on by it at a distance r and also at 
a distance /, it having changed position from ;' to / . The 
force exerted by e upon the unit quantity at these distances is 
e/r^ and e//^ (see page 33) respectively. The mean force is 
not the arithmetical average, because the force varies inversely 
with the square of the distance r. The mean force is the geo- 
metrical mean, which is the square root of the product of 
the two forces. This is \/elY^ X elr'"^ = \/e^/r"/^ = e/r/. 
To get the energy this has to be multiplied by the path 
traversed; this we have taken as the distance from r to /, 
or as r — /. 

94 



ELECTRIC ENERGY AND POWER 95 

Performing the multiplication we have 

^X(r-/) = .^'=.(l-i). (X) 

rr' rr \r rj 

If r = 00 , then energy = e/r'. (2) 

As these expressions give the energy involved in the transfer 
of unit mass, they are the expressions (i) for the numerical 
value of the potential in the case of movement from r to / and 
(2) in the case of movement from a point in space to an infinite 
distance. The latter has been defined as the value of absolute 
potential. 

It is incorrect to say that potential is equal to the energy 
required to move a unit mass from or to an infinite distance to 
or from the place of potential, because potential and energy are 
not measured in the same unit. The number of potential units 
is equal to the number of energy units as described, but there 
is no equality of potential and energy. 

Potential Drop or JJJ Drop. The current due to a definite 
e.m.f. is equal to the quotient of the e.m.f. less the RI drop in 
any part of the circuit divided by the resistance of the rest of 
the circuit. Calling the resistance of the part of the circuit 
which includes the RI drop i?i, this gives 

_ E-RJ 
R-R, ' 

The rule follows from Ohm's law. The e.m.f. expended on 
the part of resistance R^ is by Ohm's law equal to R^I. As the 
total e.m.f. expended on the system is E, the e.m.f. expended on 
the rest of the circuit is £ — R^I. The resistance of this por- 
tion, namely, the remainder of the circuit, is i? — R^. Hence by 
Ohm's law the current in the circuit is given by 

/= -^ — ^ » as above. 
K — Ki 



96 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. The fall in potential, or RI drop, in a part of a 
circuit is | volt while a current is passing due to an e.m.f. of 
3 volts in the circuit. The resistance of the remainder of the 
circuit is 2 ohms. What is the intensity of the current ? 

Solution. Subtracting the RI drop of f volt from the total 
e.m.f. of 3 volts gives the numerator of the fraction of the 

formulas 3 — — = 2^. Dividing this by the resistance of the 

4 
rest of the circuit, 2 ohms, gives 

2i -V- 2 = i8 amperes. 

Example. What is the resistance of the portion of the 
circuit in the above case in which the fall of potential of f volt 
takes place ? 

Solution. By Ohm's law R = E -r- I, and substituting, we 
have 

R = ^- -v-ii = - ohm. 
4 3 

Electric Energy. When electricity passes through a con- 
ductor it expends energy. The electric energy expended is 
converted into heat energy. The latter may be measured and 
used as the measure of electric energy. 

If a unit of electric quantity actuated by a unit of e.m.f. 
passes through a conductor, a definite amount of heat will be 
produced. If two units of quantity actuated by one unit of 
e.m.f. pass through a conductor, twice the amount will be pro- 
duced. If one unit of quantity as before but actuated by two 
units of e.m.f. passes, again twice the original amount will be 
produced. If two units of quantity actuated by two units of 
e.m.f. pass, then four times the original amount of heat will 
be produced. 

The heat due to the passage of a quantity of electricity is the 
measure of the electric energy exerted. From the statements 
of the preceding paragraph it follows that the measure of the 



ELECTRIC ENERGY AND POWER 97 

electric energy incident to the passage of electricity through a 
conductor is the product of the units of quantity by the units of 
e.m.f. To bring about the conditions of the preceding para- 
graph, the resistance of the conductor or the time employed or 
both may be changed for the different cases. 

Example. 30 C.G.S. units of quantity with 45 C.G.S. 
units of e.m.f. are expended on a conductor. How many ergs 
of energy are expended? 

Solution. 30 X 45 = i,35o ergs. 

Practical Unit of Electric Energy. The practical unit of 
electric energy is the volt-coulomb, equal to 10^ C.G.S. units of 
e.m.f. multiplied by 10-^ C.G.S. units of quantity, and there- 
fore equal to 10^ C.G.S. units of energy, or to lo^ ergs, or to the 
joule. These absolute units are in the electro-magnetic system. 

Example. 10^^ absolute units of quantity at 10 units of 
e.m.f. are how many volt-coulombs? 

Solution. 10^0 X 10 = 10" ergs. 10^^ ergs ^ lo^ = 10*, or 
10,000 volt-coulombs. 

Example. If 31 coulombs at 20 volts are expended, how 
many C.G.S. units are expended? 

Solution. 31 coulombs = 3.1 C.G.S. units. 20 volts = 20 
X 108 C.G.S. units. 3.1 X 20 X io« = 62 X lo^ C.G.S. 
units. 

Electric Powder or Activity. The rate at which electrical 
quantity passes through a conductor is current strength or 
current intensity. One unit of quantity per second is unit 
current. The product of unit current by unit e.m.f. is unit 
rate of energy, unit activity, or unit power. 

Practical Unit of Electric Power. The practical unit of 
power is the product of the practical unit of current by that of 
e.m.f. It is, as we have seen before, the volt-coulomb per 
second, the volt-ampere or watt. It is equal to 10^ C.G.S. units 
of power. 



98 ELEMENTARY ELECTRICAL CALCULATIONS 

Relations of Power to Current, Resistance, and e.m.f. 

It follows that, since the rate of energy varies with the product 
of e.m.f. by current, if either factor is constant the rate of 
energy will vary with the other one. 

Example. Calculate the value of the watt in C.G.S. units. 

Solution. The watt is the product of one volt by one 
ampere. One volt = lo^ C.G.S. units. One ampere = iq-^ 
C.G.S. units. lo^ X io~^ = lo^ C.G.S. units of rate of energy or 
of power. The C.G.S. unit of power is the erg per second. 
A rate of lo^ ergs per second is one watt. 

Example. A current of 16X10 C.G.S. units of current is 
actuated by 19 X 10^ C.G.S. units of e.m.f. Calculate the 
value in watts. 

Solution. 16 X 19 = 304; 10^ X 10 = 10^; the product of 
16 X 10 X 19 X 10' = 304 X 10^ = 3,040 X 10'^ C.G.S. units 
= 3,040 watts. 

Example. Reduce and analyze 29 watts with 7 volts into 
C.G.S. units. 

Solution. 29 -V- 7 =4.14 amperes, the current in the case. 
7 volts = 7 X 10^ C.G.S. units of e.m.f. 4.14 amperes = 4.14 
X 10-^ C.G.S. units of current. The total is 29 X Jo^ C.G.S. 
units of power. 

Example. A current of 3 amperes flows through a resist- 
ance of 3 ohms. What power is exerted? 

Solution. By Ohm's law E = IR the e.m.f. is 9 volts. 
The power is 9 X 3 =27 volt-amperes or watts. 

Example. 17 coulombs pass through a conductor of 7 ohms 
resistance in 15 seconds. Calculate the power. 

Solution. A current of 17/15 = 1.133 coulombs per second 
or amperes passes. The e.m.f. is given by Ohm's law E = IR; 
and substituting, £ = 1.133 X 7 = 7.93 volts. The power is 
therefore 7.93 X 1.133 =8.98 watts. 



ELECTRIC ENERGY AND POWER 99 

By Ohm's law E = IR. Substituting this value of E in the 
expression of power or watts EI, we have 

Power = PR, 

by which expression power may be directly calculated from the 
resistance and current intensity. 

Example. A current of 5 amperes flows through a resistance 
of 4 ohms. What is the power ? 

Solution. Substituting for P and for R their values we 
have 

Power = 25X4 = 100 watts. 

Example. A current of 2 X 10 C.G.S. units flows through 
a resistance of 4 X 10^ C.G.S. units. Calculate the power in 
watts. 

Solution. Substituting as before for P, (2 X to'^^ = 4 X lo^, 
and for i?, 4 X 10^, we have 

Power = 4 X 10^ X 4 X io» = 16 X 10" 

= 160,000 X 10^ C.G.S. units = 160,000 watts. 

By Ohm's law / = E/R. Substituting this value of / in 
the expression of power or watts EI, we have 

£2 

Power = — • 
R 

Example. Assume 200 volts to act upon a resistance of 105 
ohms. Calculate the energy absorbed per second. 

Solution. Applying the expression E'^/R, we obtain for the 
power (200)^105 = 381 watts. In a second 381 watt-seconds 
or volt-coulombs are absorbed. 

There are two expressions for electric power which include 
resistance as one of their component parts. These are PR and 

EyR. 

Assuming the current to remain constant and the resistance 
to be doubled, the first expression becomes P X 2 R, or 2 PR. 



100 ELEMENTARY ELECTRICAL CALCULATIONS 

If the resistance is trebled the same expression becomes 3 PR^ 
and so on. Therefore at constant current the electric power 
varies with the resistance. 

Assuming the e.m.f. to remain constant, and doubling and 
trebling the resistance as before, the second expression becomes 
£y2 R and ^Vs R, whence it follows that at the same e.m.f. 
the electric power varies inversely with the resistance. 

Example. The resistance of a lamp is 212 ohms. The 
conductor suppl)dng it has a resistance of 4 ohms. How 
much energy is wasted on the conductor ? 

Solution. From the law of power distribution we have 
Energy of conductor : energy of lamp : : 4 : 212. 
The energy in the conductor is 4/212 of the energy in the 

lamp and of the energy in the circuit. These 

^ 4+212 ^^ 

reduce to 1/53 and 1/54 approximately. The energy wasted 
is 1/53 that utilized in the lamp, and is 1/54 that consumed 
in the whole circuit. 

It is correct to speak of energy or of power in this relation 
because the ratios are the same for both. This problem can be 
done usually by simple inspection. 

Example. There are two lamps / and /' of resistances 
211 and 214 ohms respectively. Calculate the relative energy 
which would be absorbed by each at the same e.m.f. 

Solution. The proportion is given by the second law cited 
above, 

/:/':: 214 : 211 

or the lamp of 211 ohms resistance absorbs 2 14/2 11 of the 
energy absorbed by the lamp of 214 ohms resistance. 

Equivalents of the Watt. One watt-second is equal to 
10^ ergs. The electric energy expended in the passage of an 
electric current through a conductor is converted into heat 
energy. If the calorie is taken as equal to 4.185 X 10^ ergs, 



ELECTRIC ENERGY AND POWER lOI 

then to reduce watt-seconds to calories the number of watt- 
seconds must be divided by 4.185 or multiplied by 1/4. 185 
= 0.239, or as usually taken, 0.24. 

Example. A current of 0.5 ampere is produced in a portion 
of a circuit by an e.m.f. of no volts. What number of calories 
is absorbed by this portion per second ? 

Solution. The watts are absorbed at the rate of 0.5 X no 
= 55 per second. The calories are 55 X 0.24 = 13.2. 

Example. A current of 15.5 amperes at no volts is expended 
in an electric heater. How long will it take for it to boil a liter 
of water at 10° C. if 50 per cent of the heat is lost by waste? 

Solution. 15.5 X no = 1,705 volt-amperes or watts. 1,705 
watt-seconds multiplied by 0.24 gives 409.2 calories per second. 
A liter of water weighs 1,000 grams. To boil it, it has to be 
heated 90 degrees, making 90,000 calories that have to be 
imparted. 90,000 -v- 409.2 = 220 seconds, which is the time 
required if all the heat were utilized. But one half is lost, so 
that twice the time, or 440 seconds, is needed, or 7 minutes 
20 seconds. 

PROBLEMS. 

What current will a 120-kilowatt generator give at 120 volts, running 
at full load? Ans. 1,000 amperes. 

A current of 19 amperes actuated by 19 volts passes through a 
conductor for 8 minutes. Calculate the watt-seconds and ergs. 

Ans. 173,280 watt-seconds; 17,328 X 10^ ergs. 

WTiat is the rating of a generator that can deliver 320 amperes at 

120 volts? Ans. 38.4 kilowatts. 

Calculate the energy in 10^^ C.G.S. units of quantity at 10 C.G.S. 

units of e.m.f. Ans. io^° C.G.S. units; 10^^ joules. 

How many C.G.S. units and watts are there in 2 amperes at 9 volts? 

Ans. 18 X 10^ C.G.S. units; 18 watts. 

In 10 joules of electric energy at 8 volts calculate the quantity in 

C.G.S. units and in coulombs. 

Ans. 125 X io~^ C.G.S. units; 1.25 coulombs. 



102 ELEMENTARY ELECTRICAL CALCULATIONS 

How many joules are there in 32 C.G.S. units of quantity at 41 
C.G.S. units of e.m.f. ? Ans. 1,312 X io~^ = 0.000,131 joules. 

29 C.G.S. units of e.m.f. act on a resistance of 31 C.G.S. units. 
Calculate the power. Ans. 27.13 C.G.S. units of power. 

If a resistance of (a) 41 is in series with one of (b) 39, what per cent 
of the total power if a current in passing through them will be expended 
on each? Ans. (a) 51.25 per cent; {b) 48.75 per cent. 

A constant e.m.f. is maintained at the terminals of two lamps, one 
(a) of 120 ohms and one (b) of 115 ohms. What relative portion of 
the total energy will be absorbed by each ? 

Ans. (a) 48.9 per cent; (b) 51.1 per cent. 

What power is expended in generating 29 coulombs at 19 volts in 
II minutes? Ans. 0.835 watts. 

What is the energy developed in the above case ? Ans. 551. i joules. 
What is the value of 11 watts in C.G.S. units? 

Ans. II X 10^ = 110,000,000 C.G.S. units. 

Calculate the value of 16 X 10* C.G.S. units of current at 12 X lo^ 
C.G.S. units of e.m.f. 

Ans. 192 X 10^ C.G.S. units of power; 19.2 watts. 

If there are 31 watts of power at 32 volts, what is the current in 
amperes and C.G.S. units? 

Ans. 0.969 amperes; 969 X io~* C.G.S. units. 

171 coulombs are passed in 3 seconds through a resistance of 
8 ohms. Calculate the watts. Ans. 25,992 watts. 

The RI drop in one part of a circuit is 9 volts; the e.m.f. of the 
generator is 13 volts; the resistance of the rest of the circuit is 
II ohms. Calculate the current. Ans. 0.3636 ampere. 

What is the total resistance in the above case? Ans. 35.8 ohms. 

A number of lamps are using 51 amperes of current at no volts. 
Calculate the calories per second. Ans. 1,346.4 calories. 

A conductor is assumed of constant resistance. Compare the 

power developed in it by the passage of currents of 7 and of 17 amperes. 

Ans. 49 : 289 or as 10 : 59 (the first for the smaller current). 

How many calories per second of energy are expended on 40 lamps 
each taking | an ampere of current with a drop of no volts? 

Ans. 528 calories. 



ELECTRIC ENERGY AND POWER 16^ 

A current of 0.5 ampere is produced in a portion of a circuit by an 
e.m.f. of no volts. What number of calories is absorbed by this 
portion in a second ? Ans. 13.20 calories. 

Assume 200 volts acting upon a resistance of 105 ohms. Calculate 
the energy absorbed in one minute. Ans. 5,485.7 calories. 

A current of 9 amperes passes through a resistance of 9 ohms. 
Calculate the calories per second. Ans. 175 calories. 

How many 50-watt lamps absorb in one minute of operation the 
heat required to raise the temperature of 2 Htres of water from 15° C. 
to 100° C. ? Ans. 24 (exactly 23.6) lamps. 

A current of 5 amperes is forced through a conductor by an e.m.f. 
of 6 volts. Calculate the power. Ans. 30 watts or volt-amperes. 

A current of 5 amperes is actuated by a voltage of 4 volts. How 
many calories does it develop in one hour? Ans. 17,280 calories. 

379 calories are developed in a conductor of 3 ohms resistance. 
Determine the joules. Ans. 1,579 joules. 

If in the above example 15 seconds time was expended in the 
operation, what was the current and e.m.f.? Ans. 5.91 amperes. 

17.77 volts. 

A potential difference of 31 volts is maintained between the ends of 
a conductor of 13 ohms resistance. How many calories are developed 
in 3 minutes? Ans. 3,193.5 calories. 

25 absolute units of e.m.f. cause a current of 75 absolute units to 
flow through a conductor. How many ergs per second of energy are 
developed? Ans. 1,875 ^^gs. 

Calculate the ergs in 95 C.G.S. units of quantity actuated by 
19 C.G.S. units of e.m.f. Ans. 1,805 ^^gs. 

10^ C.G.S. units of current are maintained by 10^ C.G.S. units of 
e.m.f. What is the power? Ans. id' ergs per second = i watt. 

Between the terminals of a conductor in which 11,917,250 ergs per 
second are developed a potential difference of 3 X 10^ C.G.S. units 
of e.m.f. is maintained. What is the current? 

Ans. 0.0397 C.G.S. units. 
0.397 amperes. 

How many C.G.S. units of current and how many amperes are 
there in an electric horse-power at 75 volts? Ans. 0.995 C.G.S. units. 

9.95 amperes. 



104 ELEMENTARY ELECTRICAL CALCULATIONS 

An e.m.f. of 5 volts acts upon a resistance of 7 ohms. Calculate 
the power. Ans. 3.55 watts or volt-amperes. 

A current of 27 amperes passes through a conductor of 13 ohms resist- 
ance. An additional resistance of 17 ohms is added to the conductor, 
in series therewith, and then one of 29 ohms, also in series. Calculate 
the power expended in the three cases if the same current is main- 
tained. Ans. 9,477 watts. 

21,870 watts. 

43,011 watts. 

A uniform current is maintained through three conductors in 
parallel, each conductor being of identical resistance, and afterwards 
through the same three conductors in series. Compare the power 
expended in each case. 

Ans. I : 9, or as the square of the resistances. 

A current of 6 amperes flows through 12 lamps in parallel; each 
lamp is of 220 ohms resistance. How many watts are expended, 
and in 2 hours' running how many joules and how many ergs ? 

Ans. 660 watts. 

4,752,000 joules. 

4,752 X io^° ergs. 

How many ergs are developed per second in the operation of a 
generator delivering 21 amperes at 118 volts potential? 

Ans. 2,478 X 10' ergs per second. 

What is the power of the above generator ? 

Ans. 2,478 watts, or nearly 2I kilowatts. 

The armature of a shunt-wound dynamo has a resistance from brush 
to brush of 0.03 ohm; the e.m.f. at the brushes is no volts, with a 
current of 150 amperes in the outer circuit; the resistance of the 
shunt field is 55 ohms. What is the armature loss in watts? In a 
shunt-wound dynamo the outer circuit and shunt are in parallel. 
Give general data. 

Ans. Resistance, of outer circuit 0.733 ohm. 

Combined resistance of outer circuit and shunt 0.7235 ohm. 
Current in armature 152 amperes. 
Armature loss 693 watts. 



CHAPTER IX. 

BASES AND RELATIONS OF ELECTRIC UNITS. 

Effects of an Electric Current. — Two Systems of C.G.S. Electric Units. — 
The Basis of Practical Units. — The Basis for Measurement and 
Definition of a Current. — The Absolute C.G.S. Electro-magnetic 
Unit of Current. — The Tangent Compass. — Action of Earth's 
Field. — Angle of Divergence. — Combined action of Coil and Earth's 
Field on Magnet (See page 109). — Tangent Galvanometer For- 
mula. — Determining C.G.S. units of e.m.f. and Resistance. — The 
Absolute C.G.S. Electrostatic Unit of Quantity. — Determination 
of the E.S. Unit of Potential. — The Attracted Disk Electrometer. 
— Other Units of the Electrostatic System. —The E.M. and E.S. 
System of Units. — Equivalents of the Two Systems of Units. — 
Derivation of Practical Units from Absolute Units. — Reduction Fac- 
tor. — Problems. — Dimensions of E.M. Quantities. — Dimensions of 
Magnetic Quantity. — Dimensions of Current in E.M. System. — 
Dimensions of Electric Quantity in E.M. System. — Dimensions 
of Potential in E.M. System. — Dimensions of Resistance in E.M. 
System. — Dimensions of Capacity in E.M. System. — Dimensions 
of Electric Quantity in E.S. System. — Dimensions of Surface Den- 
sity in E.S. System. — Dimensions of Potential in E.S. System. — 
Dimensions of Capacity in E.S. System. — Dimensions of Current in 
E.S. System. — Dimensions of Resistance in E.S. System. — Dimen- 
sions of Magnetic Quantity. — Dimensions of Surface Density of 
Magnetism. — Dimensions of Magnetic Intensity. — Dimensions of 
Magnetic Potential. — Dimensions of Magnetic Power. — Dimen- 
sions of Electric Intensity in E.S. System. 

Effects of an Electric Current. Three things must co- 
exist in an electric circuit — electro-motive force, current, and 
resistance. The effects of an electric current depend upon 
the rate of electric quantity which constitutes it. If any two 
of the above are known the other can be determined by Ohm's 
law. 

Two Systems of C.G.S. Electric Units. There are two 
systems of electric units of the C.G.S. system. They are termed 
absolute units. One is the absolute system of electro-static 
units; the other is the absolute system of electro-magnetic units. 

The Basis of Practical Units. The absolute electro- 
magnetic units are usually assumed to be the basis of the practi- 

105 



lo6 ELEMENTARY ELECTRICAL CALCULATIONS 

cal units such as the volt and ampere, although the latter could 
be derived from the electrostatic units. 
The Basis for Measurement and Definition of a Current. 

If a current passes through a conductor it produces various 
effects. These effects by their intensity give a basis for measur- 
ing and defining the intensity of the current. One of these 
effects is the production of a field of force. A magnet pole is 
acted on by a field of force; it is attracted or repelled by a current. 
As the effect diminishes rapidly with the distance intervening 
between current and magnet, the effect is practically limited in 
extent to a small volume of space. The volume within which 
the effect is discernible is called a field of force. Theoretically 
a field of force may be as large as the universe. Practically 
artificial fields of force are small in volume. The field of force 
of the field magnet of a bipolar dynamo is little more in extent 
than the volume existing between its pole faces. 

The intensity of a field of force can be measured by its action 
on a magnet pole of known strength. 

A unit magnet pole is one of such strength that it will attract 
or repel another unit pole at a distance of one centimeter with a 
force of one dyne. 

The Absolute C.G.S. Electro-magnetic Unit of Current. 
The absolute C.G.S. unit of current is the following. It is 

the intensity of a current which 
• passing through a conductor i 
centimeter long bent into the arc 
of a circle of i centimeter radius will attract or repel with a force 
of I dyne a unit magnet pole placed at the center of such circle. 

The diagram shows an arc of this radius with its center 
indicated. Calling force / we have for the case described 

/=!. (I) 

Suppose that the current goes through a complete circle of 
I centimeter radius. Its action on a magnet pole at its center 




BASES AND RELATIONS OF ELECTRIC UNITS 1 07 

will be greater than in the last case in the ratio of the lengths 
of the conductors. A circle of radius i has a length or circum- 
ference of 2 Tz, The ratio of the arc of the first case to that of 
the circle of the second case is i : 2 tt. Therefore as these are 
the lengths of the conductors acting on the magnet pole, and 
as the intervening distance is the same in both cases, the force 
acting on the magnet is expressed by 

/ = 2 TT. (2) 

Example. With what force will a unit magnet pole at the 
center of a circular conductor, the circle into which it is bent 
being of i cm. radius, be acted on when a unit current passes ? 

Solution. Substituting for iz in (2) its value, 3.1416, we 
have for the force 

/ = 2 TT = 2 X 3.1416 = 6.2832 dynes. 

Suppose the wire is bent into a circle of radius ;'. The length 
of the conductor is now 2.7zr^ and as far as this change of length 
of conductor is concerned its action on the pole at its center 
varies directly with the length of the conductor acting on it. 
But the distance of the conductor from the pole is changed in 
direct ratio with r. As radiant action, which is the action 
exerted on the pole in the cases under consideration, varies 
inversely with the square of the distance, we have to express this 
action also in the next formula. The direct ratio of the length 
of the conductor and the inverse ratio of the square of the dis- 
tance give the formula 

J. 2 7rr 2 TT , . 

/ = -7- = — • (3) 

Example. Assume the radius of a conductor bent into a 
circle to be 7 cm. With what force will a unit current 
passing through it act upon a unit pole at its center ? 

Solution. Substituting in (3) the values of ;' and of it we have 

^ 2 X 3.1416 6.2832 o . J 

/ = ^-^^ — = '^ = 0.8976 dyne. 



lo8 ELEMENTARY ELECTRICAL CALCULATIONS 

Suppose that the circular conductor is wound around the 
circle a number of times n, it then ob\dously acts with n times 
the intensity of a single turn. Equation (3) then becomes 

/ = -7-' (4) 

and if the current instead of being a unit current is of strength i, 
equation (4) becomes 

/ = ——-' (5) 



whence by division 



i=-^. (6) 

2 Tin 



If the magnet pole is of strength M the force exerted upon 
ft will be M times as great as that exercised upon a unit pole. 
This is because the force exercised is a radiant force between 
two things, and such force is equal to the product of the t\vo 
forces. Equations (5) and (6) for a magnet pole of strength M 
become 

/=^-^- (7) 

2 nnM 

Example. A magnet pole of 5 units strength at the center 
of a circle of 11 cm. radius composed of 24 turns of wire is 
acted on by a force of 377 dynes. What is the intensity of 
current passing through the wire ? 

Solution. By substituting in formula (8) we obtain 

i = 377 Xii _ ^^^ ^^.^g 

6.2832 X 24 X 5 

The Tangent Compass. If in the center of such a coil a 
very short compass needle is established a tangent compass is 



BASES AND RELATIONS OF ELECTRIC UNITS 109 



produced. The compass needle is acted on by the horizontal 
component of the earth's magnetic field. The circle of the 
galvanometer is placed in the magnetic meridian approxi- 
mately north and south. A current passing through the wire 
tends to deflect the needle into a position at right angles to the 
plane of the coil. This is at right angles to the position in 
which the earth's component tends to keep it. 

Action of Earth's Field. The earth's field acts upon the 
needle with the product of the horizontal component of the 
earth's magnetism by the strength of the magnet. This prod- 
uct we may call HM^ H repre- 
senting the horizontal component 
of the earth's magnetism and M 
the force of the magnet. 

Angle of Divergence. At a 
given angle of divergence 6 from 
the magnetic meridian the lever 
arm of the earth's action on the 
magnet is equal to the product of 
the magnet's length by the sine of 
the angle of divergence. This is 
shown in the diagram. The earth 
acts upon both arms of the mag- 
net with virtual lever arms of length AB^ and the sum of the two 
is as if it acted upon the magnet with a single lever arm 
equal to the length of the magnet multiplied by sin d. 

By the same course of reasoning we find that the coil acts 
with a lever arm equal to the product of the magnet's length by 
the cosine of the angle of divergence. 

Combined Action of Coil and Earth's Field on Magnet. 
Calling the length of the magnet / we have for the earth's 




action HMl sin ^, and for the action of the coil 



nniMI cos 6 



no ELEMENTARY ELECTRICAL CALCULATIONS 

In any position of equilibrium which the needle assumes it will 
be acted on equally by these two opposing forces. They there- 
fore may be made equal to each other, or 

2 nniMl cos d -rrnri • n / x 

■ = HMl sm 6, (9) 

in which M and / eKminate each other and which by transpo- 
sition gives 

r ,T sin d rH tan d , . 

t = H — - = (10) 

2 nn cos o 2 7:n 

Tangent Galvanometer Formula. This is the formula of 
the tangent galvanometer, developed here at length to show 
how the electro-magnetic C.G.S. unit of current can be directly 
determined. It will be observed that the only constant entering 
into the formula is the horizontal component of the earth's 
magnetism. This has been determined for a great many 
places. 

Example. The radius of a coil is 30 cm., the number of 
turns is 5, the value of H is 0.1590 dyne, and the needle is 
deflected by a current passing through the coil to an angle of 
50° 11'. Calculate the current strength in C.G.S. units. 

Solution. Substituting in equation (10) we have 

_ 30 X 0.1590 ^^^ ^^, _ ^ _ ^^g ^^^ ^^.^ 

6.2832 X 5 
of current. 

Determining C.G.S. Units of e.m.f. and Resistance. 
It now remains to be shown how the C.G.S. units of e.m.f. and 
of resistance are found. 

The energy due to an electric discharge is equal to the prod- 
uct of the quantity discharged by the potential difference of the 
points or places between which the discharge takes place. The 
ergs per second of energy expended when a current of elec- 
tricity is passing through a conductor is equal to the current 
strength multiplied by the e.m.f. required to maintain the cur- 



BASES AND RELATIONS OF ELECTRIC UNITS III 

rent through the conductor in question. The energy expended 
is converted into heat energy. All units are assumed as of the 
C.G.S. system. 

By placing a conductor in a calorimeter, passing a known 
current through it, the current being taken in C.G.S. units, 
determining the ergs per second produced, the e.m.f. expended 
on the conductor is manifestly equal to the ergs divided by the 
current because the current is numerically equal to the quantity 
per second. Then having the current and e.m.f., the resistance 
of the conductor is calculated by Ohm's law. 

Example. A current of 0.114 C.G.S. units passing through 
a calorimeter wire for one minute produces 40 calories of heat. 
Calculate the e.m.f. between the terminals of the wire and the 
resistance of the wire. 

Solution. To reduce the calories to ergs they must be 
multiplied by 416.7 X 10^, whence the ergs per minute corre- 
sponding to the calories of the problem are 40 X (416.7 X 10^) 
= 1,666.7 X 10^ ergs = 1,666,700,000 ergs. Dividing the 
ergs per minute given above by 60 gives ergs per second. 
(1,666.7 Xio )-^ 60 = 278 X 10^ ergs per second. This is to be 
divided by the quantity of electricity per second, which is numer- 
ically equal to the current. (278 X lo^) -v- 0.114 = 243 X 10® 
C.G.S. units of e.m.f. The resistance of the wire is calculated 
by Ohm's law, R = E/I^ whence (243 X 10^) -4-0.1 14 = 213 
X 10^ C.G.S. units of resistance. 

I C.G.S. unit of current = 10 amperes. The current there- 
fore is 0.114 X 10 = 1. 14 ampere. 

10® C.G.S. units of e.m.f. = i volt. The e.m.f. therefore is 
(243 X 10®) -^ 10^ = 2.43 volts. 

10^ C.G.S. units of resistance = i ohm. The resistance 
therefore is (213 X 10') -^ 10® =2.13 ohms. 

This section is designed to show the relation of the absolute 
units to the practical ones and also the relation of the apparently 



112 ELEMENTARY ELECTRICAL CALCULATIONS 

abstract bases of the absolute system to the standards of the 
engineer. It is not to be taken as showing approved details of 
making these determinations, which exact the most accurate 
work and refinement in methods. 

An ammeter and a voltmeter could be standardized or cali- 
brated by the above method ; the length, diameter, and material 
of a wire of i ohm resistance could be determined, and with 
these bases the other practical units could be determined and 
standards fixed for them. 

The Absolute C.G.S. Electrostatic Unit of Quantity. 
The absolute electrostatic unit of quantity is that quantity 
which would attract or repel a similar quantity one centimeter 
distant with a force of one dyne. From this as a basis a full 
series of units of the E.S. system, as the term may be conven- 
iently abbreviated, is deduced. 

Determination of the E.S. Unit of Potential. To obtain 
an experimental basis we may start with the determination of 
potential difference. The attracted disk electrometer can be 
used for this determination. 

The Attracted Disk Electrometer. It comprises a disk 
held above and parallel to a plate, the plate being much larger 




than the disk. The apparatus is so arranged that the attraction 
between plate and disk and the distance separating them can be 



BASES AND RELATIONS OF ELECTRIC UNITS II3 



determined. In the diagram the disk is represented by the 
line S, the plate by the Hne T, and an annular plate surrounding 
the disk, as close to it as possible, is also indicated. 

By touching one of the two, disk or plate, with an excited 
conductor, a charge is imparted and they attract each other. 
Call the surface density of the charge + for one surface and — 
for the other. 

The attraction of an indefinitely large plane with a surface 
density <t for a point, in this case a unit of quantity, placed 
opposite its center, is 2 tto- (see page 270). This is the attraction 
of the plate for a unit of quantity of electricity on the disk. 
Surface density is the quantity of electricity per unit area of 
the plate or disk. 

Call the area of the disk S. Then the entire quantity on its 
surface is So-. It will be seen that a- represents the quantity on 
a unit area of the disk, because the surface densities of the 
charges on plate and disk are identical. The attraction A 
between disk and plate is therefore 

A = 2 TTO- X So- = 2 TtSo-^, (i) 

because So- is numerically equal to the number of units of quan- 
tity on the disk. 
Transposing (i) gives 

''-■As' (^) 

If a point is situated between the disk and plate, both will act, 
the one attracting and the other repelling any mass between 
them, and hence both exercising force in the same direction. 
The force at such a point will be the force exercised by one of 
the surfaces multiplied by 2, or 2 tto- X 2, = 4^0-. 

If a unit quantity is transferred from one plane to the other, 
as from disk to plate, the energy involved is numerically equal 
to the difference of potential between the two surfaces. This 



114 ELEMENTARY ELECTRICAL CALCULATIONS 

is because the product of potential difference by quantity is 
energy, and as the quantity transferred is supposed to be equal 
to unity, the potential difference and the energy have the same 
numerical value. Call the potential difference between the 
two surfaces P and the distance between the plates d. Then 
the potential difference will be equal to the force 4 ira- exercised 
on a unit of quantity multipHed by the distance of transfer d, 
giving 

P = 4 lT(Td. (3) 

Substituting for o- in this equation its value from (2) we have 



■^xv^-V'-f- 



2TtS 

The force of attraction A is measured, the area of the disk S 
and the distance d are known. Substituting for A^ S, and d 
their values, the numerical value in electrostatic units of the 
potential difference between the plate and disk is found. 

In carrying out this calculation, C.G.S. units must be used. 
The force of attraction must be expressed in dynes. 

Example. In an attracted disk electrometer, the area of the 
disk was 10 square cm. The distance from the disk to the 
plate was 0.5 cm. After imparting a charge to the disk 
the attraction was 0.092 gram. Calculate the potential differ- 
ence in electrostatic C.G.S. units. 

Solution. Substituting for the symbols of the formula (4) 
their values it becomes 



P = 0.5 4 /SttX (0.092 X981) _ ^^^ electrostatic units. 

V 10 

The part of the expression in parenthesis (0.092 X 981) 
is the force of attraction in dynes, on the assumption that the 
force of gravity was measured by an acceleration of 981 cm. 
It is equal to 90.252 dynes. 



BASES AND RELATIONS OF ELECTRIC UNITS ll^ 



Other Units of the Electrostatic System. If one such 
unit of e.m.f. were expended on maintaining a current of such 
strength that the two would expend i erg of energy per second, 
the current would be of one electrostatic absolute unit strength. 
The resistance of the conductor through which this current 
would be forced by a unit of potential difference would be one 
electrostatic unit in amount. 

From this basis the whole series of electrostatic units can be 
deduced. 

The E.M. and E.S. Systems of Units. The value of a 
current or of potential difference may be determined in absolute 
units by various methods which give it directly. Examples of 
such are given in the preceding pages, where they are employed 
simply to give the concrete idea of what these units are, of how 
they are related to the three fundamental units of time, space, 
and mass, and of how they can be determined experimentally 
by direct use of the centimeter, gram, and second. One set 
of electric units is based on the attraction of oppositely charged 
surfaces for each other, or, what is the same thing, on the repul- 
sion of similarly charged ones. This set of units is called the 
electrostatic series. Another series is based on the action of an 
electric current on a magnet pole. This series is known as the 
electro-magnetic series. Both sets of units are C.G.S. units and 
can be used for electric specification and calculation, one just as 
well as the other. 

Equivalents of the Two Systems of Units. The units 
of the two systems have values which differ from the values of 
corresponding units of the other system. The relation of the 
values has been determined by experiment, and the results are 
very accurate, yet not absolutely so. 

The E.M. absolute unit of quantity is equal to 3 X 10^® E.S. 
units. The coulomb is equal to 10-^ E.M. absolute units, and 
therefore is equal to 3 X 10^ E.S. units. 



Il6 ELEMENTARY ELECTRICAL CALCULATIONS 



Exactly the same relations exist for the current units, so that 
the ampere is equal to 3 X 10^ E.S. units. 

The E.S. unit of potential is equal to 3 X 10^^ E.M. absolute 
units. The volt is equal to 10^ E.M. absolute units, and there- 
fore is equal to — X 10-^ E=S. unit. 
3 

The E.S. unit of resistance is equal to 9 X 10^^ E.M. absolute 

units. The ohm is equal to 10^ E.M. absolute units, and there- 
fore is equal to — X 10-" E.S. unit. 
9 
The E.M. absolute unit of capacity is equal to 9 X 10^" E.S. 

units. The practical unit of capacity, the farad, is equal to 10-® 
E.M. absolute units, and therefore is equal to 9 X 10" E.S. 
units. The subject of capacity is treated of in Chapter XV. 

Derivation of Practical Units from Absolute Units. 
The practical units are derived from the absolute units by either 
of two methods, each giving an identical result. The simplest 
method is to reduce the units of the absolute series to those of the 
practical series by multiplying by defined factors. The tabic 
gives the value of the units of the practical series in terms of the 
absolute units, with the factors. The factors are all powers of 
10 or multiples thereof. 



Practical Units. 


E.M. Absolute Units. 


E.S. Absolute Units. 


Coulomb . 

Ampere 


10-^ 
lO-« 
IO« 
IO» 


3 X io« 
3 X io« 
9 X 10" 
i X 10-2 
^ X 10-" 


Farad 


Volt 


Ohm 



The relations of the absolute units are given in the following 
table. 



Quantity. . 
Current . . . 
Capacity. . 
Potential , . 
Resistance 



E.M. Abso- 
lute Units. 



E.S. Abso- 
lute Units. 



3 X io'° 
3 X io^° 
9 X io2° 
^ X io-^° 
i X 10-20 



E.S. Abso- 
lute Units. 



E.M. Abso- 
lute Units. 



i X IO-' 

X io-» 

X 10-2 

3 X ioi« 

9 X 1020 



BASES AND RELATIONS OF ELECTRIC UNITS II/ 

Example. Calculate the equivalent of 5 volts in electrostatic 
units (absolute). 

Solution. A volt is equal to 10^ absolute E.M. units; 5 volts 
therefore are equal to 5 X 10^ absolute E.M. units. For the 
equivalent electrostatic units this must be divided by 3 X lo^*'. 
Then 

5 X io« -V- 3 X io^° = - X 10-2 = 0.0166 E.S. units. 
3 

Or by direct process, using the equivalent of the table, 5 X J X 

10-2 = 0.0166 E.S. units. 

Example. What is the value of the volt in absolute electro- 
static units ? 

Solution. From the above calculation it follows that the value 
of the volt is 0.0166 -4- 5 = 0.0033 E.S. unit. 

Or by the table, i X J X io~^ = 0.0033 E.S. unit. 

Example. What is the value of 28.3 E.S. absolute units in 
volts ? 

Solution. As the E.M. unit is equal to the E.S. unit divided 
by 3 X 10^*^, the reverse rule holds and the E.S. unit is equal to 
the E.M. unit multiplied by 3 X lo^'^. This gives 

2S.S X 3 X 10^0 =, g^^ X jo9 absolute E.M. units. 

To reduce this to volts we must divide by the equivalent of the 
volt in absolute E.M. units, which is 10*. 

(849 X 10^) -^ 10* = 8,490 volts. 

Or by the table, 28.3 X 3 X 10^ = 8,490 volts. 

Example. An e.m.f. of 120 volts produces a current of 
5 amperes. Reduce these quantities to E.S. absolute units; 
apply Ohm's law to the result, thus obtaining the value of the 
resistance in the same system of units; reduce the resistance 
thus calculated to ohms and test the correctness of the opera- 
tion by Ohm's law directly applied. 



Il8 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. The absolute value of 120 volts in the E.M. 
system is 120 X 10^ = 12 X 10^. Applying the equivalent to 
obtain the value of this quantity in absolute E.S. units we 
have 

12 X 10^ -^ (3 X lo^o) = 4 X 10"' = e.m.f. E.S. 

5 amperes are equal to 5 X io~^ absolute E.M. units. Apply- 
ing the equivalent we find 

5 X 10-' X (3 X 10'^) = 15 X 10^ = current in E.S. units. 

Applying Ohm's law to the E.S. units thus determined gives 

4 X 10-^ -^ (15 X 10^) = 266 X IO-13 = resistance E.S. 

To test the correctness of the operation proceed thus: The 
original data give the resistance by Ohm's law as £// = 120 
-^ 5 = 24 ohms. 

24 ohms = 24 X 10® absolute E.M. units. Applying the 
equivalent we find 

24 X 10^ -T- (9 X io^°) = 266 X 10^^, as found by the former 
operation. This goes to prove the correctness of the work. 
This example is only given as an exercise. The results are 
better obtained by the use of the tables. 

The practical units of the E.M. system can be directly de- 
rived from the dimensional formulas of units by changing the 
units of length and mass. 

Let the unit of mass be io~" gram and let the unit of length 
be 10^ cm., the unit of time being the second, as in the 
absolute system. If these are introduced into the E.M. dimen- 
sional formulas of units the resulting units will be of the practi- 
cal system, volts, amperes, ohms, and the regular engineering 
units. 

Example. Calculate the value of the ampere in absolute 
electro- magnetic units. 



BASES AND RELATIONS OF ELECTRIC UNITS II9 

Solution. The dimensions of the unit are M^L^T-^. Sub- 
stituting for these symbols their values as given above we have 
for the ampere 

lo-V X 10^ X i-^ = 10-^ C.G.S. units! 

Ten amperes are equal to one absolute electro-magnetic unit. 

Example. Calculate the value of the ohm and volt as 
above. 

Solution. For the ohm the dimensions are LT-^, and for 
the volt M^L^T-^. Substituting as above gives 

Ohm = lo^ X i-^ = 10^ C.G.S. units. 
Volt = io-°2° X 10'^' X I-' = 10' C.G.S. units. 

Reduction Factor. The value of the reduction factor of 
the two systems of C.G.S. units, which is also the numerical 
value of the velocity of light in centimeters, is about 3 X 10*®, 
a fact applying to the electro- magnetic theory of Hght. 

The reduction factor has been determined with close approxi- 
mation to the exact figure by several investigators with results 
which are reasonably concordant. The following are some of 
the results reached by recent observers. 

1883, J. J. Thomson 2 .963 X lo^" 

1889, Lord Kelvin 3 .004 X io^° 

1890, J. J. Thomson and G. F. C. Searle 2 .9955 X io^° 

, O. Lodge and Glazebroke 3 .009 X 10^'' 

Example. Calculate the dimensions of the reduction factor 
to reduce the E.S. unit of capacity to the E.M. unit. 

Solution. The dimensions of the E.S. unit are L; of the 
E.M. unit, Lr^T^. To find the factor the second must be 
divided by the first, thus : 

Lr^ T^ -^ L = L-^ T^j which is the reciprocal of velocity 
squared. 

Example, Make the same calculation for current units. 



120 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. The dimensions of the E.S. unit are ikf^L^T-^; of 
the E.M. unit, M^L^T-^. Dividing as above we find 
MLlr-2 -f- MiLir-^=Lr-S which ^j.^ the dimensions of 
velocity. 

Dimensions of E.M. Quantities. Two magnet poles of 
equal strength and of opposite polarity attract each other in 
direct proportion to the square of the quantity of magnetism in 
one of the poles and in inverse proportion to the square of the 
distance between them. 

Dimensions of Magnetic Quantity. Call the magnetism 
of each pole Q, and let the distance separating them be denoted 
by L. The attraction of the poles is a force, and we have seen 
that the dimensions of a force are MLT-^. The force in the 
particular case is, as just indicated, Q^L-^. Putting these two 
equal to each other, as they are the same, we have 

Q^L-^ = MLT-\ whence Q" = ML^T-^ and 
Q=M^L^T-\ 

which are the dimensions of magnetic quantity. Care must be 
taken not to confuse magnetic quantity with electric quantity. 
They are totally distinct. 

Dimensions of Current in E.M. System. Let / denote 
the strength of a current passing through a conductor of length L 
and bent into an arc of a circle of radius L as explained. 
Let a magnet pole representing a quantity of magnetism Q be 
at the center of the circle of which the conductor is an arc. 
The force exercised between the current and magnet pole will 
follow the laws of radiant action, with the additional law 
that the action will vary directly with the length of the con- 
ductor. The action will be therefore the product of the current 
strength by the length of the conductor by the quantity of 
magnetism, the whole divided by the square of the distance 
separating the two loci of force, namely, the conductor and the 



BASES AND RELATIONS OF ELECTRIC UNITS 121 

magnet pole. This gives the expression for the action between 
the two, 

ILQ -^ L^ = IQL-\ 

As this action is force, it can be put equal to the dimensions 
of force, or 

IQL-^ =MLT-\ 

Q is magnetic quantity, whose dimensions have just been 
determined and are M^L^T~'^. Substituting these for Q in 
the last equation gives 

/ {M^L^T-^) L-i = / (M^L^r-i) - MLT-\ whence 

which are the dimensions of electric current in the electro- 
magnetic system. 
Dimensions of Electric Quantity in E.M. System. The 

quantity of electricity passed by a current in any given time 
is evidently equal to the product of the current by the time 
during which it is passing, as it is clear that a given current 
will pass twice as great a quantity in two minutes that it will in 
a single minute. Multiplying the dimensions of current by the 
dimensions of time T gives the dimensions of electric quantity, 
M^L^T-^ X r, which reduces to 

These dimensions are quite different from those of magnetic 
quantity. 

Dimensions of Potential in E.M. System. The product 
of e.m.f . by electric quantity is energy. Therefore if the dimen- 
sions of energy MUT-^ are divided by those of electric quan- 
tity M^L^ the dimensions of e.m.f. will be given, or 

Dimensions of Resistance in E.M. System. According 



122 ELEMENTARY ELECTRICAL CALCULATIONS 

to Ohm's law resistance is equal to the quotient of e.m.f . divided 
by current strength. Carrying out this division with the dimen- 
sions as just determined gives the dimensions of resistance, thus : 

These are also the dimensions of velocity or rate of motion. 

Dimensions of Capacity in E.M. System. The capacity 
of a surface, as of a condenser, is defined as determined by the 
quantity of electricity required to raise its potential a given 
amount. Its dimensions are therefore equal to the dimensions 
of quantity M^L^ divided by the dimensions of e.m.f. M^L^T^'^j 
or if W -J- M^L^ r-2 = Z-i T^. 

The dimensions of electric energy, so called, are the same 
as those of every other form of energy, ML^T-^. 

The same applies to electric power, whose dimensions are 
MUT-\ 

Dimensions of Electric Quantity in E.S. System. A 
quantity of electricity acts upon another equal quantity at a 
distance d with force, following the laws of radiant action, 
and such force varies accordingly directly with the square of 
the quantity q and inversely with the square of the distance d. 
Thus calling force / we have / = (f/d"^ = q^d-^, whence 
^2 = ^2y ^^^ q = d \/f = df^. Substituting for d its dimen- 
sion L and for / its dimensions, those of force, MLT~^, gives 



q=LX "^MLT-^ = M^LiT-\ 

the dimensions of electric quantity. 

Dimensions of Surface Density in E.S. System. Sur- 
face density is the quantity per unit area; D is the expression 
for unit area. Dividing the dimensions of quantity as just deter- 
mined, M^L^T-^, by V gives 

the dimensions of surface density. 



BASES AND RELATIONS OF ELECTRIC UNITS 1 23 

Dimensions of Potential in E.S. System. Electrostatic 
potential, which is e.m.f. in the electrostatic system, is energy 
ML2r-2 divided by quantity M^LiT-\ or e.m.f. = ML^r-^ --- 
M^L^T-^ =MiLiT-\ 

Dimensions of Capacity in E.S. System. Capacity is 
quantity of electricity per unit of e.m.f., as in the electro-mag- 
netic system. This gives 

k = M^L^T-' -r MhL^T-"^ = L. 

Thus electrostatic capacity can be expressed as a length. 
Dimensions of Current in E.S. System. Current is 
quantity per unit time, giving 

i = M^L^T-"- ^ T = MiL^T-\ 

Dimensions of Electric Intensity in E.S. System. Electric 
intensity is force per unit of electric quantity (M^Lf 7^^), or 

h=MLT-^ (force) -v- M^LIT^^ = MiL~hr-\ 

Dimensions of Resistance in E.S. System. Resistance 
is given by Ohm's law as e.m.f. divided by current, or 

Dimensions of Magnetic Quantity. Quantity of mag- 
netism is deduced from the law of the tangent galvanometer, 
as it may be referred to for convenience, given on page no. 
A quantity q^ of magnetism at the center of a circular current 
of intensity i, of radius r, and of length / is acted on by a 
force as explained in the section referred to, which force is 
given by the expression / = ilq' -^ r^, whence 

and as f = MLT-\ r^ = L^, i = M^L^T-\ and / =L, we 
find by substituting these values 

q' = (MLT-'' X L2) -j- (M^L^T-^ X L) = M^L^, 

the dimensions of quantity of magnetism. 



124 ELEMENTARY ELECTRICAL CALCULATIONS 

Dimensions of Surface Density of Magnetism. Surface 
density of magnetism is the quotient of the above expression 
divided by unit area L ^, or 

Dimensions of Magnetic Intensity. Magnetic intensity is 
force per unit of quantity, L^M^. This gives 

Dimensions of Magnetic Potential. Magnetic potential 
is based on the following facts as regards the determination 
of its dimensions. The product of magnetic intensity by 
quantity gives energy, just as the product of a volt by a cou- 
lomb does in practical units. . If the quantity is a unit it is 
evident that the energy will vary with the potential. In other 
words the energy when the quantity is of unit value will be 
numerically equal to the potential. This is a statement 
widely different from the one so often used, that for unit 
quantity the potential is " equal to " or " is work or energy." 
The equality is only numerical and only exists for unit value 
of quantity. This gives 

e" = ML^T-^ (energy) -7- M^L^ (quantity) 
= M^L^ 7^^ (magnetic potential). 

Dimensions of Magnetic Power. Magnetic power is the 
product of magnetic density by thickness of shell, or 

^ = ilf^L"! (density) XL = M^LTh. 

PROBLEMS. 

Express 17 amperes in the E.S. system. Ans. 51 X 10® E.S. units. 
Express 39 volts in the E.S. system. 

Ans. 0.13 E.S. unit, or 13 X iQ-^ E.S. units. 

Express 29 ohms in the E.S. system. 

Ans. 3S X 10-" E.S. unit, or 322 X 10-" E.S. units. 



BASES AND RELATIONS OF ELECTRIC UNITS 1 25 

_ ^, , , I volt . , „ ^ 

Express Ohm s law = i ampere m the E.S. system. 

I ohm 

Ans. ^ X 10-2 ^ i X 10-" = 3 X io» E.S. units. 

If 13 X io~^ E.S. unit of potential acts upon a resistance of 322 

X io~*^ E.S. units, what will the current be in E.S. absolute, in E.M. 

absolute, and in practical units? Ans. 4,037 X 10^ E.S. absolute. 

0.1346 E.M. absolute. 

1.346 ampere. 

Express 15 coulombs in E.S. and E.M. absolute units. 

Ans. 45 X 10' E.S. units. 

1.5 E.M. units. 

What is the ratio of the E.S. to E.M. units of quantity from the 

above? Ans. E.S. = E.M. X 3 X io^°. 

Give the calculation for the watt in E.S. units. 

Ans. 3 X io« X I X 10-2 = 10^ E.S. units. 

Give the same calculation for E.M. units. 

Ans. 10^ X 10-^ = 10^ E.M. units. 

How many volts are there in 390 E.S. units of potential? 

Ans. 1.30 volts. 

How many amperes are there in 277 E.M. units of current? 

Ans. 27.7 amperes. 



CHAPTER X. 

THERMO-ELECTRICITY. 

Emissivity. — Heating of a Conductor by a Current. — Cross Section of 
a Conductor to be Heated to a given Degree by a Given Current. — 
Thermo-electric Couple. — Electro-motive Force of a Thermo-electric 
Couple. — Neutral Temperature. — Temperature and Electro-motive 
Force of Thermo-electric Couple. — Thermo-electric Tables. — Peltier 
Effect. — Absolute Temperature. — Thomson Effect. 

Emissivity. If a body is heated above the temperature of 
its surroundings it parts with its heat. The process is termed 
emissivity. Unit emissivity is the quantity of heat lost per 
second per square centimeter of surface per degree of difference 
between its temperature and that of its surroundings. 

It is measured in ergs or calories or other heat unit. The 
C.G.S. unit of heat is the erg. The practical unit is the therm, 
calorie, or gram-degree, which is the heat required to raise one 
gram of water one degree C. 

Heating of a Conductor by a Current. A number of 
formulas for calculation of the temperature imparted to a con- 
ductor by a current passing through it have been proposed. 
From the nature of the case these formulas can only be approxi- 
mate. 

By radiation and convection about 0.00,025 calorie is lost 
per second by an unpoHshed metalHc surface of one square 
centimeter in the air for each degree C. that it is heated above 
the air. If therefore the calories expended on a conductor by 
the passage of a current are determined, and if the wire is assumed 
to have attained its fixed temperature, these calories will be 
equal to those lost by radiation and convection. The quotient 
of these calories divided by the area of the surface of the con- 
ductor in square centimeters will give the calories per square 
centimeter which are expended on the conductor and which 

126 



THERMO-ELECTRICITY 12/ 

are in turn lost by it. The quotient of these calories divided by 
0.00,025 is the temperature in degrees C. which the wire will 
attain. The result is only approximate. 

Example. A current of 10 amperes passes through a wire 
one centimeter in circumference. The wire has a resistance of 
0.3 ohm per 100 meters. How many degrees will its temperature 
be increased ? 

Solution. The watts per second due to the passage of the 
current are given by the expression PR = 100 X 0.3 = 30. 
The surface area of 100 meters of the wire are 10,000 square 
centimeters. The calories expended on 100 meters are 30 X 
0.24 = 7.2. The calories per square centimeter are the quotient 
of 7.2 by the area of the wire, or 7.2 X io~^ = 0.00,072 calorie. 
Dividing 0.00,072 by 0.00,025 gives 2.88° C.,the degrees C. above 
the temperature of the air to which the wire would be heated by 
such a current. 

A formula for the rise in temperature in degrees C. of a bare 
copper wire with a current passing through it in still air is 
given below. The square of the amperes is divided by the cube 
of the diameter in mils, and the quotient is multiplied by 50,000. 
The product is the increase in temperature. 

P 

Temperature C. = — X 50,000. 

0/ 

Example. A copper wire 125 mils in diameter is conducting 
a current of 10 amperes. How much will it be heated above the 
temperature of the surrounding air, the wire being of bare copper? 

Solution. Substituting in the formula the above values gives 

io2 
Temperature C. = — j X 50,000 = 2.5° C. 

In the above example the thickness of the wire is the same as 
in the preceding example. The results are reasonably close for 
approximate methods. 



128 ELEMENTARY ELECTRICAL CALCULATIONS 

Cross Section of a Conductor to be Heated to a Given 
Degree by a Given Current. The diameter of a wire of a 
material of known specific resistance which a stated current will 
heat to a stated degree of heat above that of the surrounding air 
can be calculated approximately by the appHcation of the emis- 
sivity factor. 

The following discussion leads to results in centimeters. 

Specific resistance is given in microhms in the tables. The 
resistance in ohms of a wire of diameter d and whose cross- 
sectional area is therefore nd?/^ is given in the equation 

R = sp. res. X io"'-f- — = sp. res. X lo"' X ~ (i) 
4 Tza^ 

The heat expended on a conductor in the passage of a current 
is PR watt-seconds or joules per second. Substituting for R 
its value from (i) gives 

Heat= PX sp. res. X io~^ X — ^ watt-seconds per second. (2) 

A calorie is equal to 4.16 watt-seconds. If the second member 
of (2) is divided by 4.16 it becomes 

„ ^ P Xsp.res. X io~^ X 4 , • j / s 

Heat = — ; calories per second. (3) 

4.16 X ^^^ ^ 

The surface area of one centimeter of the wire is 7td. If the 
second member of (3) is divided by tt d^ the result will be the heat 
per square centimeter of surface area. Accepting 4,000 as the 
emissivity factor, the heat emitted per square centimeter at the 
temperature t is ^/4,ooo. If the wire attains a constant temper- 
ature, the heat expended and that emitted are the same. There- 
fore if the second member of (3) is divided by nd it can be 
equated with ^/4,ooo, giving 

t _ PXsp. res. X 10"^ X 4 .v 

4,000 4.16 X T^d^ ' ^^^ 



THERMO-ELECTRICITY 129 

Multiplying (4) throughout by — ^ gives 

, P X sp. res. X io~° X 4 X 4>ooQ 
~ tXTt'X 4.16 



7^ X sp. res. X 0.00039 
~ t 



(5) 



and d = ^/._2iiPj:£^X^^2239 . (6) 



= r 



To calculate the diameter of a cylindrical conductor which 
will attain a certain temperature in the air in passing a certain 
current, multiply the square of the current by the specific resist- 
ance and by the factor 0.00039 and divide the result by the tem- 
perature. The cube root of the product will be the diameter of 
the conductor in centimeters. 

Example. Calculate the diameter of a lead wire to melt with 
a current of 7.2 amperes. The melting point of lead is taken as 
335° C., and its specific resistance is 19.85 (microhms per cubic 
centimeter). 

Solution. By formulas (5) and (6) the cube of the diameter is 
obtained by substitution, gi\ing 

^3 ^ 5ij 4 X 19.85 X 0.00039 ^ 
335 



d =Vo.ooii9 = 0.1058 cm. 

Recurring to equation (4), the term j^/nH^ is the reciprocal of 
the product of the surface area of the unit length of a con- 
ductor by the cross-sectional area of the same. This product 
is 7r(?/4 X Tid, and its reciprocal is ^/Tt^d^, as it appears in the 
second member of equation (4). This equation is restricted 
to a circular conductor. By substituting for r.d?/^ and for nd 
the proper values the equation may be made to apply to con- 
ductors of other cross sections. 



•130 ELEMENTARY ELECTRICAL CALCULATIONS 

Suppose a strip is to be cut from a sheet of metal of thickness a, 
and it is to be calculated how wide the strip shall be to attain 
a given temperature with a given current. Call the width of 
the strip na, the factor n being the unknown quantity to be 
determined. The cross-sectional area of the strip will be 
na y, a — na^. The surface area of the unitary length will 
be 2 na -{- 2a = 2 a {n + i). The product of these two quantities 
is 2a^ (n^ + n). As this is the product of the surface area of 
the unit length of the conductor by its cross-sectional area, it is 

obvious that it can be substituted for in equation (4). 

4 

This substitution is performed by division by it in place of 

division by . giving 

4 



(7) 



t ^ P X sp. res. X lO"* 
4,000 4.16 X 2a(n^ + n) 

Multiplying by v ^ :^ gives 

L 

2 , P X sp. res . X i o~^ X 4,000 

4.16X^X2^3 

_ P X s p. res. X 0.0,004,808 .^v 

txa^ <^^ 

The value of n is obtained from this equation by substituting 
the constants given in the problem. The width of the strip is 
the product of its thickness by n, or na. 

Example. A strip of lead to be melted by a current of 7.2 
amperes is to be cut from a sheet 0.0 1 cm. thick. The 
specific resistance of lead is taken as 19.85 and its melting point 
as 335° C. Calculate the width of the strip. 

Solution. Substituting these values in equation (8) gives 

n^ + n = (7-^)' X^9-8.^X 0.0 004,808 ^^^ ggg 
335 X (o.oi)* 



I 



THERMO-ELECTRICITY 1 3 1 

Solving this we find 

n = 38, approximately. 

The width of the strip is the product of the thickness by this 
factor. It is therefore 

38 X o.oi = 0.38 cm. 

Thermo-electric Couple. If two conductors of different 
materials, such as iron and copper, are joined at both ends and 
separated at the middle so as to form a species of ring, and if 
the junctions are maintained at different temperatures with 
proper relation to what is called the neutral temperature a 
current of electricity will be produced, continuing as long as the 
difference of temperature is maintained. It is sufficient if one 
pair of ends is in actual contact, the pieces then separating and 
having their other ends connected by a wire. The arrange- 
ment described constitutes a thermo-electric couple. The 
intensity of the current produced depends upon the e.m.f. and 
upon the resistance of the circuit. 

Electro-motive Force of a Thermo-electric Couple. 
The e.m.f. produced by a thermo-electric couple depends upon 
the difference of temperature of the ends referred to the neutral 
temperature of the particular couple. If the junction of a 
couple is heated above the neutral temperature a certain num- 
ber of degrees and if the other ends are the same number of 
degrees below the neutral temperature there will be no e.m.f. 
produced. Different couples have widely different neutral 
temperatures. For each couple there is a fixed neutral tem- 
perature. 

Neutral Temperature. When the mean temperature of 
the ends of the couple is below the neutral temperature the 
current flows in one direction; when the mean temperature is 
above the neutral temperature the current flows in the other 
direction. 



132 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. The neutral temperature of a thermo-electric 
couple of two particular metals is 275° C. One end is heated to 
a temperature of 300° C, the other end to a temperature of 
250° C. What will the result be ? 

Solution. The average temperature is (250° + 300°) -^ 2 = 
275°. Thus the mean or average temperature of the ends is the 
neutral temperature of the couple. No potential difference will 
be maintained and consequently no current will pass. The 
result as far as the electric state of things is concerned will be 
zero. 

Example. The junction of a couple is heated to a temper- 
ature of 675° F., the other ends are kept at a temperature of 
55° F. and no current nor difference of e.m.f. can be produced. 
What is the neutral point of the combination ? 

Solution. The average temperature is (675° + 55°) -r- 2 
= 365°. As no current is produced at this mean temperature of 
the ends, or, what is the same in effect, as no e.m.f. is produced 
at this mean temperature, the neutral temperature of the couple 
is 365° F. 

Temperature and Electro-motive Force of Thermo- 
electric Couple. The e.m.f. produced by a thermo-electric 
couple depends on the difference of temperature maintained 
between the opposite ends of the couple. The e.m.f. produced 
by various combinations of elements of thermo-electric couples, 
such as an iron-copper couple, a bismuth-antimony couple, 
and so on, is different for different couples. 

To increase the e.m.f. of a thermo-electric battery the couples 
must be arranged in series. The laws given for the arrange- 
ment of battery cells apply in general to thermo-electric couples. 

Thermo-electric Tables. A thermo-electric table is a 
table of constants by which the e.m.f. of different couples can 
be calculated. In some such tables no allowance is made for 
the neutral temperature. It is obvious that such a table is only 



THERMO-ELECTRICITY 1 3 3 

accurate for one mean temperature. This is not a serious 
defect, because the mean temperature of a thermo-electric 
battery is not subject to a wide variation. 

THERMO-ELECTRIC TABLE FOR A NEUTRAL •TEM- 
PERATURE OF ABOUT 20° C. 

(Jenkin's " Electricity and Magnetism " compiled from Matthiessen's experiments 
reduced to C.G.S. units.) 

Bismuth, pressed commercial wire — 9,700 

Bismuth, pure pressed wire — 8,900 

Bismuth, crystal, axial — 6,500 

Bismuth, crystal, equatorial — 4,500 

Cobalt , — 2,200 

German silver — 1,175 

Mercury — 41.8 

Lead o 

Tin + 10 

Copper, commercial + 10 

Platinum + 90 

Gold + 120 

Antimony, pressed wire + 280 

Silver, pure hard + 300 

Zinc, pure pressed + 370 

Copper, electrolytic + 380 

Antimony, pressed commercial wire + 600 

Arsenic + 1.356 

Iron, pianoforte wire -|- 1,750 

Antimony, crystal, axial + 2,260 

Antimony, crystal, equatorial + 2,640 

Phosphorus, red + 2,970 

Tellurium + 50,200 

Selenium + 80,700 



To calculate the e.m.f. of a combination subtract algebraically 
the number opposite one element of the couple from the num- 
ber opposite the other element. The result is the e.m.f. in 
C.G.S.. units corresponding to a difference of temperature of 
i°C. betsveen the ends of the couple. The polarity of the 
e.m.f. is such as to produce a current from the lower to the 
upper metal through the hotter end or junction. 

Example. Calculate the e.m.f. between lead and cobalt at 
a mean temperature of 20° C. with a temperature difference of 
1° C. between the junctions or ends. 



134 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. The numbers opposite the metals in the table 
are o and —2,200. Subtracting gives 2,200 C.G.S. units of 
e.m.f. As 10^ C.G.S. units = i volt, this is 2,200 -r- 10^ = 2,200 
X 10-^ = 0.000,022 volt. 

Example. Assume that the elements in the above example 
are joined in contact at one end, while a wire connects their 
free ends, and that the heat is appHed to the junction. In which 
direction will the current go? 

Solution. The heated junction is the hottest. Cobalt is the 
lower metal in the table. Therefore the polarity of the e.m.f. 
is such as to produce a current from the cobalt bar through the 
hotter junction to the lead. 

Example. If the difference of temperature in the above 
case were 7° C, what would the voltage be ? 

Solution. 0.000,022 X 7 = 0.000,154 volt. 

Example. A thermo-electric battery of German silver and 
iron elements at a mean temperature of 20° C. with a temper- 
ature difference between the ends of the couples of 121° C. 
and with the couples arranged in series gives an e.m.f. of 1.33 
volts. Calculate the number of couples in the series. 

Solution. Iron, + 1,756, and German silver, —1,175, ^^e 
to be subtracted algebraically. Changing the sign of German 
silver and adding we obtain 1,756 + 1,175 = 2,931. This is 
the e.m.f. of a single couple in C.G.S. units at i°C. difference 
of temperature between the ends. At 121° C. the difference 
is 2,931 X 121 = 354,651 C.G.S. units. 1.33 volts (the e.m.f. 
of the battery in volts) = 1.33 X 10^ = 133,000,000+ C.G.S. 
units. Dividing the e.m.f. of the battery in C.G.S. units by 
the e.m.f. of a single couple in the same units gives the number 
of couples. 133,000,000 -^ 354,651 = 375, which is the num- 
ber of couples required to produce 1.33 volts at the difference of 
temperature specified in the problem. 

The e.m.f.'s of the above calculation could have been expressed 



THERMO-ELECTRICITY 1 3 5 

in volts and the operation could have been done in these units. 
It is merely a question of decimal place. Thus : 

354,651 C.G.S. units = 0.003,546,510 volt and 
1.33 -7- 0.003,546,510 = 375 couples, as before. 

In these operations it is immaterial which number has its 
sign changed, becoming thereby the subtrahend. It is only 
necessary to change the sign of one of the numbers and to add 
the two algebraically. The position in the table determines 
the polarity. 

The table given below takes in a considerable range of mean 
temperature and is a more satisfactory one to work with than 
is the last. It is based upon the work of Professor Tait (Trans. 
Royal Soc, Edinburgh, Vol. XXVII, 1873). The table is taken 
from Everett's *' C.G.S. System of Units." 

THERMO-ELECTRIC HEIGHTS AT t° C IN C.G.S. UNITS. 

Iron + i>734 — 4-87/ 

Steel +i>i39 - 3 28 f 

Alloy, platinum, 95; iridium, 5 + 622 — o .53 ^ 

Alloy, platinum, 90; iridium, 10 + 596 — i .34 ^ 

Alloy, platinum, 85; iridium, 15 + 709 — o .63 < 

Soft platinum — 61 — 1.10/ 

Alloy, platinum and nickel + 544 — 1.10/ 

Hard platinum + 260 — 0.75^. 

Magnesium + 244 — o .95 ^ 

Cadmium 4- 266 + 4 .29 / 

Zinc + 234 + 2 .40 ; 

Silver 4- 214 4- i .50 f 

Gold 4- 283 4- 1.02^ 

Copper 4- 136 + 0.95; 

Lead o 

Tin - 43 + 0-55^ 

Aluminum — 77 4- o .39 / 

Palladium — 625 — 3 .59 ^ 

German silver — 1,207 — 5 .12 f 

Nickel to 175° C — 2,204 — 5 -12 f 

Nickel 250° to 310° C — 8,449 + 24 .1 t 

Nickel from 340° C — 307 — 5.12^ 

To use the table the mean temperature of the couple in degrees 
C. is substituted for t 



136 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. Calculate the e.m.f. of an iron-German silver 
couple, the temperatures of the junctions being 0° C. and 
100° C. 

Solution. The mean temperature is the average tempera- 
ture of the ends, or (100 + o) -7- 2 = 50° C. This is to be 
substituted for t of the table, to obtain the values of the ele- 
ments o: the couple, thus: 

Value of iron Ij734 — 4 .87 ^ 

Value of German silver — 1,027 — 5.12^ 

Algebraic difference 2,941 + o .25 / 

Substituting]; for t its value, 50, in the third expression, we have 
e.mi. (at i°C.) of couple = 2,941 + (0.25 X 50) = 2,953.5 
C.G.S. units. The difference of temperature, by the conditions 
of the problem, is 100 degrees; we therefore have to multiply 
the e.m.f. for i°C., as calculated above, by 100 to obtain the 
e.m.f. for 100 degrees difference, or 

e.m.f. (at 100° C. difference of temperature) = 2,953.5 X 
100 = 295,350 C.G.S. units. As 10^ C.G.S. units are equal to 
I volt, we have 

295,350 H- 108 = 295,350 X 10"^ = 0.0,0295 volt. 

The value of t could have been substituted in the expressions 
for the values of the elements of the couple, thus: 

Value of iron . . . 1,734 - - 4 "7 ^ = i'734 - (4 -87 X 50) = 1,490.5 
Value of German 

silver - 1,207 - 5 .12 / = - 1,207 - (5 .12 X S^) = — ^>4^3 

Algebraic difference • 2,953.5 

giving the same result as before, the value of the couple for 
i°C. 

Example. Calculate the neutral point of the above couple. 

Solution. The neutral point is the value of t when there is 
no e.m.f., or when 2,491 + 0.25 t = o, whence t = — 9,964° C. 



THERMO-ELECtRiCITY 1 3^ 

There is therefore no neutral point for this couple, the calcu- 
lated one being below the absolute zero. 

Example. Calculate the neutral point of iron and copper. 

Solution. Proceeding as before we find from the table : 

Value of iron i>734 — 4 -87 ^ 

Value of copper 136 +0.95^ 

Algebraic difference i>598 — 5 .82 ^ 

Making the algebraic difference zero gives to t sl value which 
is that of the neutral point. Thus: 

1,598 — 5.82 t = o, whence t = 275° C. 

Peltier Effect. If a current passes through a circuit of 
different metals there is a production of heat or a reduction of 
heat at the junctions of different metals independent of the 
ordinary heating effect of the current. This is called the Peltier 
effect. The value of the Peltier effect is expressed in ergs per 
second per C.G.S. unit current (10 amperes). 

Absolute Temperature. To determine the Peltier effect 
multiply the difference of thermo-electric heights at the junction 
in question by the absolute temperature of the junction, all in 
degrees C. The absolute temperature is obtained by adding to 
the temperature C. the number 273, which is the absolute tem- 
perature of the centigrade zero. 

Example. What is the absolute temperature of boiling 
water ? 

Solution. The temperature C. of boiling water is 100°. 
To obtain the absolute temperature add 273, or 

100 + 273 = 373° absolute temperature C. 

Example. Assume a C.G.S. unit current to pass through a 
junction of iron and copper maintained at a temperature of 
100° C. What is the value of the Peltier effect ? 



138 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. The thermo-electric height of iron at 100° C. 
is 1,734 — (4.87 X 100) = 1,247. That of copper is 135 4- 
(0.95 X 100) = 230. The difference of thermo-electric heights 
is 1,247 ~ 230 = 1,017. The absolute temperature at the 
junction is 100 + 273 =373. Then following the rule we 
have 

1,017 X 373 = 379,341 ergs per second for a C.G.S. unit 
current. 

If the current flows from iron to copper through the heated 
junction it will be a current from the upper to the lower ele- 
ment. This produces a heating effect. If the current is from 
the lower to the upper element, in this case from copper to 
iron through the hot junction, the effect w^ill be to cool the 
junction. In other words, if the direction of the current is 
such as to supplement the thermo-electric current, it will cool 
the junction, otherwise the reverse will hold. 

Thomson Efifect. If a conductor of one material through- 
out is heated in places so that some parts are hotter than 
others, it will show thermo-electric action, except in the case of 
lead. The action is called the Thomson effect. 

Referring to the table (page 135), metals with a minus 
sign prefixed to their temperature coefficients, in the second 
column of figures, are affected as follows: An electric current 
passing from a hotter to a cooler portion reduces the tempera- 
ture of the cooler part. Passing from a cooler to a hotter por- 
tion it heats the conductor in the hotter part. This applies 
to iron and all other metals with negative signs in the table. 

Metals with a positive sign prefixed are affected in the re- 
verse way. A current passing from hot to cool heats the cool 
portion. Passing from cool to hot it lowers the temperature of 
the hot portion. Copper is one of the metals subject to this 
action. 

The general law is that in the case of metals with a negative 



THERMO-ELECTRICITY 1 39 

sign an electric current tends to increase any local differences 
of temperature already existing. In the case of metals with 
positive signs an electric current tends to equalize temperature 
differences. 

If portions of a conductor differ in temperature, the product 
of that difference by the thermo-electric temperature coefficient 
of the metal as given in the table gives the thermo-electric 
difference between the portions in question. The temperature 
difference is to be in degrees C, and the result will be in C.G.S. 
units. 

Example. The ends of an iron wire differ by 100° C. in 
temperature. Calculate the thermo-electric difference between 
the ends. 

Solution. From the table the thermo-electric coefficient of 
iron is found to be— 4.87. Then, as the sign has no effect 
upon the multiplication except as indicating the polarity of the 
e.m.f., we have 

100 X 4.87 = 487 C.G.S. units of e.m.f., or 0.000,004,87 volt. 

As the coefficient has a negative sign, the current tends to 
increase any existing differences of temperature. 

The value of the Thomson effect in ergs per second per C.G.S. 
unit current (10 amperes) is found by multiplying the thermo- 
electric difference, as just calculated in the case of iron for 
instance, by the sum of one-half the temperature difference in 
degrees C. and 273. This reduces the temperature to the 
absolute scale C. 

Example. Calculate the Thomson effect for the iron wire of 
the last example. 

Solution. The temperature difference is 100° C. ; half of 
this is 50. Following the rule we have 

487 X (273 -f 50) = 157,301 ergs per second for a current of 
10 amperes (i C.G.S. unit). 

If the current goes from cold iron to hot iron, this number of 



140 ELEMENTARY ELECTRICAL CALCULATIONS 

ergs of electric energy are converted into heat energy; if the 
current goes the other way, the same number of ergs of heat 
energy are converted into electric energy. 

Example. Let a copper wire be unequally heated ioo° C. 
as in the last example. Calculate the Thomson effect. 

Solution. The thermo-electric effect will be 0.95 X 100 = 95 
C.G.S. units. For the ergs per second we have 

95 X (273 + 50) = 30,685 ergs per second for a lo-ampere 
current. 

Example. Make the same calculation for lead. 

Solution. As the temperature coefficient for lead is o, there 
is no Thomson effect for lead. 

PROBLEMS. 

Calculate the e.m.f. between iron and lead per degree C. at 0° C. 

Ans. 1,734 X io~^ volt. 

Calculate the e.m.f. of iron and copper, the junctions being kept 
at 0° C. and 100° C, Ans. 130,700 X 10-® volt. 

A thermo-electric battery of S3 bismuth (— 9,700) and antimony 
(2,250) couples shows 1.25 volts at the mean temperature of 19° C. 
What is the difference of temperature between the two faces ? 

Ans. 317° C. 

A current of 11 amperes is passed through a wire of 0.7 cm. circum- 
ference and of 0.6 ohm resistance per 100 meters. How much will its 
temperature be increased ? Ans. 10° C. 

What will be the rise in temperature in a copper wire of no mils 
diameter carrying a current of 20 amperes? Ans. 15° C. 

Calculate the diameter of a lead wire to melt with a current of 3.1 
amperes. Ans. 0.0606 cm. 

Taking the melting point of copper as 1,127° C. and its specific 
resistance as 1.652, calculate the thickness of a wire that will melt at 
21 amperes current. Ans. 0.0632 cm. 

A copper strip is 0.02 cm. thick. How wide should it be to melt with 
an 1 8-ampere current? Ans. 0.097+ cm. 



THERMO-ELECTRICITY 1 4 1 

How wide should the same copper strip be for ten times as great a 

current? Ans. 1.0586 cm. (more than ten times the first width). 

Calculate the e.m.f. of bismuth (— 9,700) and platinum per degree C. 

at 20° mean temperature C. Ans. 0.000,097,9 volt. 

Make the same calculation for German silver and copper. 

Ans. 0.000,015,55 volt. 

How many couples of the above would be required for a volt at 100° C. 

difference of temperature ? Ans. 608 couples. 

Calculate by table on page 190 the e.m.f. per degree C. difference, 

of German silver and copper at 20° C. mean temperature. 

Ans. 0.000,014,64 volt. 
What is the e.m.f. per degree C. for a mean temperature of 315° C. ? 

Ans. 0.000,032,55 volt. 
At 600° C. difference of temperature what would be the number of 
couples of the combination of the last problem to a volt ? 

Ans. 53 couples. 
What is the neutral point of the above combination? Ans. — 22i°C. 
What is the neutral point of iron and zinc? Ans. 206° C. 

What is the neutral point of aluminimi and palladiimi? 

Ans. - 138° C. 

Calculate the e.m.f. per degree C. of aluminum and nickel at a mean 

temperature of 300° C. ? Ans. 0.000,012,69 "^^^t. 

What would the voltage be at 600° C. difference of temperature, at 

above mean temperature, and how many couples would be required for 

a volt? Ans. 0.007,614 volt; 132 couples. 

Calculate the Thomson effect in a copper wire, with a difference of 

200° C. in temperature of different sections of its length. 

Ans. 0.000,001,9 volt; 70,870 ergs per second. 
Make the same calculation for German silver with 75° C. difference 
of temperature. Ans. 0.000,003,84 volt; 119,232 ergs per second. 

Make the same calculation for silver with 98° C. difference in tem- 
perature. Ans. 0.000,001,47 volt; 47,334 ergs per sec. 
What is the Peltier effect between German silver and iron at 100° C. 
Ans. 1,106,318 ergs per second for a C.G.S. unit current. 
Calculate the Peltier effect for zinc and tin at 190° C. 

Ans. 290,995 ergs per second for a C.G.S. unit current. 



CHAPTER XI. 

ELECTRO-CHEMISTRY. 

Chemical Composition. — Chemical Saturation. — Chemical Equivalents 
and Atomic Weights. — Electric Decomposition or Electrolysis. — 
Relation of Chemical Equivalents to Electrolysis. — Electro-chemical 
Equivalents. — Thermo-electro Chemistry. — Calculation of Elec- 
tro-motive Force of a Voltaic Couple. — Problems. 

Chemical Composition. The atomic weights of the ele- 
ments indicate the proportions in which the elements combine, 
subject to Avogadro's law. Thus the atomic weight of hydrogen 
being i and that of oxygen being i6, it follows that in their 
combination with each other the ratio of i : i6 or some 
multiple thereof must obtain. The water molecule contains 
two atoms of hydrogen and one atom of oxygen. The ratio 
follows from the atomic weights; it is 2 : 16. The atomic weight 
of the hydrogen is multiplied by 2. This example illustrates a 
fundamental law of chemistry, the law of multiple proportions. 

The determination of the atomic weights is one of the most 
difficult problems of the analytic branch of chemistry, and 
atomic weights are subject to constant revision as new 
determinations are made. 

Example. The formula of sulphuric acid is H2SO4. Calcu- 
late the proportions of the constituent elements. 

Solution. Multiplying the atomic weight of each element by 
the number of the atoms of it which are in the mglecule we 
have 

H = 2 X I = 2 . S = I X 32 = 32 •= 4 X 16 = 64. 

This gives the ratio 

H : S : O : : 2 : 32 : 64. 

Example. The formula of sodium sulphate is NagSO^. 
Calculate the proportions of the constituent elements. 

142 



ELECTRO-CHEMISTRY 1 43 

Solution. Sodium : sulphur : oxygen :: 46 : 32 : 64. 

Example. Calculate the proportions of the constituent 
elements of silver nitrate, AgNOg (N= 14). 

Solution. Silver : nitrogen : oxygen : : 107. i : 14 : 48. 

Chemical Saturation. When two elements combine so as 
to satisfy their chemical affinities they are said to saturate each 
other. If a single atom of one element saturates a single atom 
of another element, the two are said to have the same valency. 
Thus the formula of copper oxide is CuO. It is completely 
saturated; therefore both copper and oxygen are of the same 
valency, or atomicity, as it is also called. 

Valency. If other ratios than the unitary obtain in a simple 
saturated molecule, it indicates that the valencies of the con- 
stitutent elements differ one from the other. The formula of 
the water molecule is HgO; therefore oxygen has a valency 
twice as great as that of hydrogen. 

Elements of a valency of one are called monads; those 
of a valency of two, dyads; of three, triads; of four, tetrads; 
of five, pentads; of six, hexads; of seven, heptads; of eight, 
octads. 

Example. The valency of hydrogen is i. What is the 
valency of oxygen, the water molecule being a saturated one 
and its formula being H2O? 

Solution. As it requires two atoms of the monad hydrogen 
to saturate the one atom of oxygen, oxygen must be a dyad. 

Example. The formula of sulphur trioxide is SO3. What is 
the valency of sulphur ? 

Solution. As one atom of sulphur saturates three atoms of 
the dyad oxygen, the valency of sulphur is 3 X 2 = 6 ; sulphur 
is a hexad. 

The statements given above are rather illustrative than 
exhaustive, as other considerations apply in many cases in 
chemistry. It is also to be noted that in practical work the 



144 ELEMENTARY ELECTRICAL CALCULATIONS 

less important decimals are omitted from the atomic weights in 
calculations. 

Chemical Equivalents and Atomic Weights. The chem- 
ical equivalent of an element is the quotient of its atomic weight 
divided by its valency. It expresses the simplest ratio of the 
constituent elements of any simple saturated molecule of two 
elements, a binary molecule as it is termed. 

The atomic weight of oxygen is i6; it is a dyad; its chemical 
equivalent is therefore i6 -^ 2 =8. The atomic weight of 
hydrogen is i; it is a monad; therefore its chemical equivalent 
is I. The molecule of water has the formula HgO. Substituting 
for the constituent symbols their chemical equivalents gives 
I : 8 as the ratio of the oxygen and hydrogen in water. 

Example. Calculate the chemical equivalents of sulphur. 

Solution. Sulphur in some compounds is a hexad. Its 
atomic weight is 32; its chemical equivalent is 32 -^ 6 =5.3. 
In other compounds, in SO2 for example, it is a tetrad. The 
chemical equivalent of tetrad sulphur is 32 -^ 4 = 8. Some- 
times, as in H2S, it is a dyad, when its chemical equivalent 
is 32 -^ 2 = 16. 

Electric Decomposition or Electrolysis. When a current 
of electricity is passed through a compound so as to electrolyze 
it, a definite amount of the compound is decomposed by a 
definite amount of electricity. For various reasons some com- 
pounds cannot be electrolyzed. The statement of the necessary 
requirements for electrolysis is outside the field of this book. 
Assuming the possibility of the separation of any element from 
its compounds by electrolysis, the quantity of an element actually 
or potentially separable by a definite quantity of electricity will 
be referred to as the weight or quantity '' corresponding to " the 
quantity of electricity in question. In the same way the quan- 
tities of electricity " corresponding to " given weights of the 
elements and of their compounds will be referred to. 



I 



ELECTRO-CHEMISTRY 



45 



Chlorine 

Chromium 

Copper (cuprous) 
Copper (cupric) . . 

Gold 

Hydrogen 

Iron (ferrous) .... 

Iron (ferric) 

Lead 

Mercury ( — ous). 
Mercury ( — ic) . . 

Nickel 

Oxygen 

Platinum 

Silver 

Sodium 

Sulphur. 

Sulphur 

Sulphur 

Tin (stannous) . . . 
Tin (stannic) .... 
Zinc 



Atomic 
Weight. 



35-18 

5^-7 
63.1 



195-7 

I 

55-5 



205-35 
198.5 



58.3 
15.88 

193-3 
107. II 
22.88 
31.82 



64.9 



Val- 
ency. 



Chemical 
Equiv- 
alent. 



35-18 
8.62 
63.1 

31-55 
65-23 

I 

27-75 
18.50 
102.67 
198.5 
99-25 
29.15 

7-94 

32 . 22 

107. II 

22.88 

15-91 

7-95 

5-30 

59-05 

29.52 

32-45 



Electro- 
chemical 
Equivalent 
for C.G.S. 

Unit. 



003672 
000900 
006586 
003293 
006809 
,00010438 
,002897 
.001931 
.010717 
.020719 
.010360 
• 003043 
.000829 
•003363 
.01118 
.002388 
.001661 
.000831 
■000553 
.006164 
. 003082 
.003387 



Recipro- 
cal. 



272.3 
1,111 .4 

151-83 
303 . 66 
146.87 
9,580.4 

345-24 
517.86 

93-31 
48.26 

96.53 

328.66 

1,206.6 

297-34 

89 -45 

418.72 

602.16 

1,204.32 

1,806.48 

162.24 

324.48 

295.24 



It is not necessary for ordinary work to use more than one 
decimal, and in many cases none need be expressed. Thus 
107 may often be used as the atomic weight of silver, and 16 
is almost always used as the atomic weight of oxygen. 



146 ELEMENTARY ELECTRICAL CALCULATIONS 

Relation of Chemical Equivalents to Electrolysis. Chem- 
ical equivalents are the relative weights of elements correspond- 
ing to any definite quantity of electricity. 

In some molecules atoms of the same element partly saturate 
each other. Chemical equivalents do not apply to such. In 
the molecule CjHg the tetrad carbon elements partly saturate 
each other, so that the carbon is virtually a triad. In the molecule 
CgHi4 it has a virtual valency of 2 J. 

The quantities of elements which a definite quantity of elec- 
tricity will precipitate are proportional to their chemical equiv- 
alents. 

Example. A certain quantity of electricity would precipitate 
119 grams of silver. What weight of copper would the same 
quantity precipitate ? 

Solution. The chemical equivalents of silver and of copper 
are respectively 107 and 31.6. We then have the proportion 

107. 1 : 31.6 : : 119 : x = 35.1 grams. 

Example. A current of electricity is passed through two 
solutions, one a solution of silver, the other a solution of an 
unknown salt. From the first solution 129 grams of silver are 
precipitated; from the other solution 35.1 grams of an unknown 
metal are precipitated. What was the metal ? 

Solution. The ratio of the metals precipitated is the ratio of 
their chemical equivalents. This gives the proportion 

129 : 35.1 : : 107. i (the chemical equivalent of silver) : x 
(the chemical equivalent of the unknown metal) = 29.1. 

By referring to the table this is seen to be the chemical equiva- 
lent of nickel, which therefore is the metal of the unknown 
solution. 

Electro-chemical Equivalents. The electro-chemical equiv- 
alent of an element is the weight in grams corresponding to, 




ELECTRO-CHEMISTRY 1 47 

or which would be precipitated by, one C.G.S. unit of electricity. 
A table could be made out for any desired unit. 

The electro-chemical equivalents of the elements are pro- 
portional to their chemical equivalents. 

Example. The electro-chemical equivalent of hydrogen is 
0.000,104,38. What is the electro-chemical equivalent of 
sodium ? 

Solution. The chemical equivalent of sodium is 23 and that 
of hydrogen is i ; therefore its electro-chemical equivalent is the 
product of 0.000,104,38 X 23 = 0.002,401. 

Example. The electro-chemical equivalent of silver is 
0.01118. Calculate the electro-chemical equivalent of zinc by 
direct proportion. 

Solution. The chemical equivalents of silver and of zinc are 
respectively 107. i and 32.45, retaining the decimals. As the 
electro-chemical equivalents are proportional to the chemical 
equivalents the proportion follows: 

107. 1 : 32.45 : : 0.01118 : x =0.003,387, which is the elec- 
tro-chemical equivalent of zinc. 

Example. Calculate the electro-chemical equivalent of a 
dyad element whose atomic weight is 118.1. 

Solution. The chemical equivalent is the quotient of the 
atomic weight divided by the valency, or 118.1 -^ 2 = 59.05. 
The electro-chemical equivalent is the product of the electro- 
chemical equivalent of hydrogen, 0.000,104,38, by this number. 

0.000,104,38 X 59 = 0.006,164, which is the electro-chemical 
equivalent of stannous tin. 

The reciprocal of the electro-chemical equivalent is the number 
of units of electricity required to precipitate one gram of the 
substance or element in question. Thus to precipitate one gram 
of hydrogen i -v- 0.000,104,38 = 9,580.4 C.G.S. units of elec- 
tricity are required. 



148 ELEMENTARY ELECTRICAL CALCULATIONS 

To reduce this to coulombs multiply by 10. To precipitate one 
gram of hydrogen 9,580.4 coulombs are required. 

To obtain the electro-chemical equivalent of any element the 
electro-chemical equivalent of hydrogen may be multiplied by 
the chemical equivalent of the element. To obtain the reciprocal 
of any element, divide the reciprocal of hydrogen, 9,580.4, by 
the chemical equivalent of the element. 

Example. Calculate the electro-chemical equivalent and its 
reciprocal for the metal lead. 

Solution. The electro-chemical equivalent of hydrogen is 
0.000,104,38. Multiplying this by the chemical equivalent of 
lead, 102.67, gives 0.010,717, the electro-chemical equivalent of 
lead. Dividing 9,580.4 by 102.67 gives 93.31, the reciprocal 
for lead. 

Example. In one hour a current of electricity precipitates 
3.740 grams of silver. What is the strength of the current ? 

Solution. There are 3,600 seconds in an hour. If we divide 
3.740 by 3,600, the quotient, 0.001,039, is the quantity of silver 
precipitated in one second. One ampere precipitates 0.001,118 
gram of silver in one second. Therefore 0.001,039 -^ 0.001,118 
= 0.93 amperes is the strength of the current. 

Example. How many grams of silver are deposited in an hour 
by one ampere ? 

Solution. One ampere-hour is equal to 3,600 coulombs. One 
C.G.S. unit deposits 0.01118 gram, and one coulomb deposits 
0.001,118 gram, because one C.G.S. unit is equal to 10 coulombs. 
0.001,118 X 3,600 = 4.0248 grams, which is the weight of silver 
deposited in an hour by one ampere, or by 10 amperes in A 
hour. 

Example. For how long must a current of 7 amperes be 
maintained to precipitate 13 grams of silver ? 

Solution. From the last column we find that to precipitate 
one gram of silver 89.45 C.G.S. units are required. These are 



ELECTRO-CHEMISTRY 1 49 

equal to 894.5 coulombs. To precipitate 13 grams 894.5 X 13 
== 11,628.5 coulombs are required. A current rate of 7 amperes 
is 7 coulombs per second, so that 11,628.5 -^ "j =1,661.2 seconds, 
or 27 minutes 41 seconds would be required. 

Thermo-electric Chemistry. Let H be the heat in calories 
due to the combination of one gram of any element with another. 
Assume that Q coulombs have acted; then if we denote the 
electro-chemical equivalent of the element by z, the weight of 
the element involved in the reaction will be Qz grams, and the 
heat will be QzH calories. One calorie is equal to 0.424 kilo- 
gram-meters of energy. Therefore the kilogram-meters of the 

reaction will be ^ xr / ^ 

0.424 QzH ... (i) 

The energy of the reaction can be expressed by direct reference 
to electrical energy. If Q coulombs are involved in a reaction, 
the energy is equal to Q X £ joules, E denoting the volts of the 
same reaction. One joule is equal to 1/9.8 1 kilogram-meters 
Therefore the kilogram-meters of the reaction will be 

pr • • • ^'^ 

Equating (i) and (2) we have 

2| = 0.424 QzH ... (3) 

and solving (3) we find 

E = 4.16 zH . . . (4) 

Calculation of Electro-motive Force of a Voltaic Couple. 

This expression gives the value of the e.m.f. of a reaction in 
terms of the electro-chemical equivalent and of the calories of 
heat per gram. The latter quantity has been determined for a 
great many chemical reactions. 

Example. What e.m.f. is required to decompose water ? 

Solution. Take the value 34,450 calories as the heat due to 



I50 ELEMENTARY ELECTRICAL CALCULATIONS 

the oxidation of one gram of hydrogen, producing water. The 
electro-chemical equivalent of hydrogen is, for coulomb notation, 
0.000,010,438. Introducing these values in (4) gives 

E — 4.16 X 0.000,010,438 X 34,450 = 1-495 volts. 

These calculations are not in exact accord with direct experi- 
ment, as the heat coefficient is not always exact, and subsidiary 
reactions may exist which are not taken into account in the 
calculation. 

In the action of a battery, chemical decomposition, operating 
to reduce the e.m.f., sometimes has a place. Then in calculating 
the e.m.f. of such a battery the e.m.f. of decomposition must be 
subtracted from the e.m.f. of combination to get the e.m.f. of 
the battery. 

Example. Calculate the e.m.f. of the Daniell battery. 

Solution. In the Daniell couple zinc is dissolved in sulphuric 
acid. In the solution of one gram of zinc in sulphuric acid 1,670 
calories are set free. The electro-chemical equivalent of zinc is, 
for coulomb notation, 0.000,338,7. Substituting in (4) we have 

E = 4.16 X 0.000,338,7 X 1,670 = 2.353 volts. 

This is the voltage due to the chemical combination of the 
couple. There is also a decomposition, that of the copper sul- 
phate, from which copper is deposited. In the solution of copper 
in sulphuric acid 881 calories are set free or are absorbed in 
its separation, the two actions being strictly reciprocal. This 
reciprocal relation obtains in all chemical reactions. The 
electro-chemical equivalent of copper is 0.000,329,3 . Substituting 
in (4) we have 

E = 4.16 X 0.000,329,3 X 881 = 1.207 volts. 

As copper is separated in this reaction heat is absorbed and 
the voltage is reduced. The net voltage of the couple is therefore 

2.353 "~ i«207 = 1. 146 volts. 



ELECTRO-CHEMISTRY 151 

By actual test the voltage of the Daniell couple is found to be 
1.079 volts, a discrepancy of 0.067 volts. 

In e.m.f. or polarization calculations such as those Just 
illustrated the calculation may be based on any of the constituents 
of the reaction or on the final product. The result of the calcu- 
lation will be the same. To prove this we may use the reaction 
of hydrogen and oxygen in the formation of water. 

In the reaction in question i part of hydrogen unites with 8 
parts of oxygen to form 9 parts of water. The calories due to 
the reaction are 34,450 for i gram of hydrogen, as we have seen. 
This then is the heat due to the combination of 8 gramas of oxygen 
with hydrogen, or due to the formation of 9 grams of water. 
Therefore for i gram of oxygen the heat is 34,450 -i- 8 = 4,306 
calories, and for i gram of water it is 34,450 -^ 9 = 3,828 
calories. In applying the formula these are the quantities which 
have to be multiplied by the electro-chemical equivalents of 
oxygen or water, as the case may be. But the electro-chemical 
equivalent of oxygen is equal to that of hydrogen multiplied by 
8, and that of water is equal to that of hydrogen multiplied by 9. 
The result of the multiplication of the formula is therefore the 
same in all the three cases. The electro-chemical equivalent of 
oxygen is 0.000,082,9 ; that of water is 0.000,093,3. The formulas 
for the oxygen and water basis are obtained by substituting the 
respective factors in (4), giving 

For oxygen, £ = 4.16 X 0.000,083 X 4,306 = 1.487 volts, 
For water, £ = 4.16 X 0.000,093,34 X 3,828 = 1.487 volts, 
which are the same as those obtained on the hydrogen basis. 

The quantity of electricity required for depositing the number 
of grams of an element equal to its chemical equivalent is the 
same for all elements. It is 9,580.4 C.G.S. units, or 95,804 
coulombs. Thus this quantity of electricity will deposit 107. i 
grams of silver, 29.15 grams of nickel, 65.23 grams of gold, 
29.52 grams of tin, and so on. 



152 ELEMENTARY ELECTRICAL CALCULATIONS 

The e.m.f. of a couple can be determined from this factor. 
The energy of a cell is expressible in two ways. It can be 
expressed in energy units, such as ergs, or in compound electric 
units, each one the product of a unit of e.m.f. by a unit of quan- 
tity. These two must be equal to each other. If the quantity 
of electricity is known, it is obvious that the quotient of the 
energy unit divided by the quantity unit will give the e,m.f. 

Example. Calculate the e.m.f. of the Daniell couple, using 
the above factor. 

Solution. 9580.4 C.G.S. units of electricity will precipitate 
32.45 grams of zinc. The calories of heat due to the solution 
of I gram of zinc we have taken as 1,670. The calories due 
to 32.45 (the chemical equivalent of zinc) grams of zinc are 
equal to 32.45 X 1,670 = 54,191.5 calories. 

The calories for the corresponding figure for copper are given 
by the product of 31.55 (the chemical equivalent of copper) 
by 881 (the calories corresponding to i gram of copper). Thus 
31.55 X 881 = 27,796 calories. 

The total calories of the couple are equal to 54,191.5 — 
27,796 = 26,395.5 calories. The copper which is deposited 
involves the reduction of the heat units. 

To reduce calories to C.G.S. units or ergs they must be 
multiplied by 4.2 X 10^. 

26,395.5 X 4.2 X 10^ = 11,086 X 10^ ergs. 

These ergs correspond to a definite quantity of electric energy. 
Of this quantity one constituent is known. This constituent is 
9,580.4 C.G.S. units of quantity. If therefore the ergs are 
divided by this figure, the quotient will be the other constituent 
or factor. This other factor is e.m.f. expressed in C.G.S. units. 

(11,086 X 108) -T- 9,580.4 = 1.16 X 108 C.G.S. units = 1. 16 volts. 

Example. Calculate the e.m.f. of the decomposition of water. 
Solution. The chemical equivalent of hydrogen being i, the 



ELECTRO-CHEMISTRY 1 5 3 

calories per equivalent are the same as those per gram, namely, 
34,450. The operation is then the same as the last, except that 
the calories per gram of equivalent do not have to be calculated. 

34,450 X 4-2 X lo^ = 14,469 X 10^ ergs. 
(14,469 X 10^) -4- 9,580.4 = 1. 51 X 10^ C.G.S. units = 1.51 volts. 

PROBLEMS. 

Express the operation of calculating the electro-chemical equivalent 
of cupric copper. Ans. 31.55 X 0.000,104,38 = 0.003,293. 

With what factor must the electro-chemical equivalent of any 
element be multiplied to give grams per hour? 

Ans. As I ampere =-y*^ C.G.S. unit, the factor is 3,600^-10 = 360. 

If 0.0397 gram of silver is precipitated by a current, what is the 
quantity of electricity in C.G.S. units and in coulombs? 

^^•^- 3-55 C.G.S. units; 35.5 coulombs. 
Using the reciprocal of the electro-chemical equivalent of gold, 
146.87, express the operation of determining the quantity of electricity 
required to precipitate 0.0781 gram of gold. 

Ans. 0.0781 X 146.87 = 11.47 C.G.S. units; 114. 7 coulombs. 

How long will it take 3.4 amperes to precipitate 1.751 grams of 
cupric copper? Ans. 26 minutes 4 seconds. 

A current precipitates from a solution of copper sulphate 0.115 
gram of copper; from a second solution in series with the first the 
same current precipitates 0.23 gram of copper. What is the second 
solution? Ans. A solution of cuprous copper. 

The same current precipitates from one solution 0.071 gram of 
silver and from another solution in series with the first 0.0658 gram of 
another metal. What is the other metal? Ans. Mercuric mercury. 

What weight of gold will 3.75 amperes deposit in 5 hours 21 
minutes? Ans. 49.18 grams; 758.94 grains. 

What proportion will give the relative weights of cupric copper 
and gold precipitated by the same current, taking gold as unity? 

Ans. 652 : 316 : : I : rv = 0.484. For each grain of gold 0.484 grain 
of copper will be deposited. 

What weight of lead will be deposited by 3.9 amperes in 10 seconds? 

Ans. 0.042 gram; 0.648 grain. 



154 ELEMENTARY ELECTRICAL CALCULATIONS 

A current deposits 1.75 grams of silver in i hour. Calculate its 
strength. Ans. 0.4348 ampere. 

The atomic weight of a pentad element is 14.01. What is its 
chemical equivalent? Ans. 2.802. 

A current passing through a solution deposits 0.00231 gram of 
cupric copper each second. The e.m.f. expended is 1.22 volts. Cal- 
culate the power or activity. Ans. 8.55 watts. 

A current passing through an electrolytic voltameter and then 
through a calorimeter deposits 16. i grams of silver per hour and devel- 
ops 6,912 calories. What e.m.f. was expended in the calorimeter? 

Ans. 2 volts. 

The current in the above case is increased so as to precipitate 18 
grams of silver per hour. What will be the effect on the calorimeter ? 

Ans. It will show 7,728 calories. 

A current of electricity is passed through two solutions, one of silver 
and one of cupric copper. After a time it deposits 5 grams of silver. 
How much copper will be deposited? Ans. 1.473 grams. 



CHAPTER XII. 

FIELDS OF FORCE. 

Fields of Force. — The Unit Field. — Intensity of Field. — Polarity. — 
Quantity of Field, — Lines of Force per Square Centimeter and per 
Square Inch. — Kapp's Unit. — Fields of Uniform and of Varying 
Strength. — Radiant Fields of Force. — Reciprocal Action in Fields 
of Force. — Induction of E.M.F. and Current. 

Fields of Force. A field of force exists in any region in 
which, without physical contact, force is exercised upon a mass 
or. other physical quantity. Thus when a magnet exercises 
force upon a piece of iron, attracting it without any contact 
bet^^-een the two, a magnetic field of force is shown to exist in the 
space between the two, because magnet and iron are drawn 
together by a force exerted in the space in question without any 
physical contact or connection betw^een the two. The force 
exerted in a field of force may be one of repulsion, as when two 
similar magnet poles repel each other. 

It is known that every mass attracts every other mass without 
physical contact or connection. This fact shows the existence 
of a field of force, which is called the field of force of gravity or 
of gravitation. The electrician is principally concerned with 
the electro-magnetic and electrostatic fields of force, especially 
with the electro-magnetic field. 

Like other physical quantities there are two things to be 
specified or measured in a field of force — its total value and its 
intensity. The unit of measurement for both of these is usually 
the line of force. For intensity the lines of force per unit area 
are to be specified, as loo lines to the square inch or 15.5 lines 
to the square centimeter. 

The Unit Field. Intensity of Field. A unit field of force 
is one which acts with unit force on a unit quantity. Thus a 

155 




156 ELEMENTARY ELECTRICAL CALCULATIONS 

unit electro-magnetic field acts on a unit quantity of magnetism 
with unit force, which force is a dyne. A field is said to vary in 
strength or intensity as it acts with more or less force on any 
given quantity. A unit field of force has i line of force to the 
square centimeter. 

Another unit of strength of the electro-magnetic field is the 
gauss, a unit field being of one gauss intensity. A gauss indicates 
a strength of one line of force to the square centimeter. 

Example. A magnet pole of 1.339 units strength is acted on 
by an electro-magnetic field with a force of 39.7 dynes. What 
is the strength of field ? 

Solution. A unit pole would be acted on in the same field with 
a force of 39.7 -5- 1.339 — 29.65 dynes. The field is of 29.65 
gausses intensity or strength or of 29.65 lines of force to the 
square centimeter. 

Example. With what force will a 12-gauss field act on a 
magnet pole of 5.5 units strength ? 

Solution. With 12 X 5.5 =66 dynes. 

Polarity. A field acts to attract or repel quantities subject 
to its action. The direction in which these actions are exerted 
is the polarity of the field. If a magnet needle is pivoted in a 
magnetic field it will take a position coinciding with the polarity 
of the field at the place it occupies and lying parallel with the 
lines of force. If a conductor is said to be moved across a field 
at right angles to the fines of force, it would be moved at right 
angles to the longitudinal axis of the magnet. A weight is 
attracted to the earth; the direction of its fall indicates the 
polarity of the earth's gravitation. 

Quantity of Field. The intensity of a field multiplied by 
its cross-sectional area gives a quantity of field. Thus if a field is 
of 5 units strength, an area of 3 square cm. will include what may 
be called a quantity of 15 units. The calculations referring to 
electric fields are generally done in lines of force, and as a unit 



I 



FIELDS OF FORCE 1 5/ 

field is one of i line to the square centimeter the above quantity- 
is simply a total of 15 lines of force. Sometimes a line of force 
is called a weber. 

Example. A field of force of 19 lines to the square centimeter 
acts upon a unit magnet pole. What is the force in dynes 
exerted upon the pole ? 

Solution. As a line of force is equal to i dyne acting upon 
unit quantity, and as the magnet pole is a unit pole, the answer 
is 19 dynes. 

Example. An electro-magnet attracts a magnet pole of 31 
units strength with a force of 51 grams. What is the strength 
of field? 

Solution. 51 grams are equal to 51 X 981 = $0,031 dynes. 
This is the action of the field upon a quantity of magnetism of 
31 units. Its action upon a unit quantity would be ^j as 
great, or would be equal to 50,031 -i- 31 = 1,614 lines of force 
to the square centimeter. 

Example. Expressing the strength of the earth's gravity in 
gravitational lines of force, how many lines of force to the square 
centimeter would it contain? 

Solution. The earth's gravitational field acts upon a unit 
mass, I gram, with a force of 981 d)mes (approximately). Its 
strength is therefore 981 lines of force to the square centimeter. 
Lines of Force per Square Centimeter and per Square 
Inch. The lines of force of a field are often referred to the 
square inch. A square inch is equal to 6.45 square cm.; there- 
fore the number of lines of force per square centimeter is 
reduced to lines per square inch by multiplying by 6.45. A 
square centimeter is equal to 0.155 square inch; so that the 
number of lines of force per square inch is reduced to lines per 
square centimeter by multiplying by 0.155. 

Example. The field of a dynamo has 5,000 lines per square 
centimeter. Calculate the lines per square inch. 



158 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. Multiplying as above we have 5,000 X 6.45 = 
32,250 lines per square inch. 

Example. Reduce 121,300 lines per square inch to lines per 
square centimeter. 

Solution. Multiplying by the reduction factor gives 121,300 
X 0.155 = 18,801 lines per square centimeter. 

Kapp's Unit. The Kapp line of force is equal to 6,000 C.G.S. 
Hues of force to the square inch and consequently to 930 lines 
of force to the square centimeter. It is a unit which is but little 
used. 

Example. How many Kapp lines are there in a field 10 
square inches in section with 3,906 lines per square inch? 

Solution. The total number of Hues is 3,906 X 10 = 
39,060. Dividing this by the equivalent, 6,000, we obtain 6.51 
Kapp lines. 

Example. Calculate the Kapp lines in a field of 27 square cm. 
section with 2,750 lines to the square centimeter. 

Solution. Proceeding as above, but using the other equiva- 
lent, we have 2,750 X 27= 74,250 as the total number of lines, 
and dividing by 930, the equivalent for square centimeters, gives 
79.8 Kapp lines. 

Fields of Uniform and of Varying Strength. A field of 
force may be of uniform strength in all its parts or ma/ vary 
according to any law. The Hues of force of a uniform field are 
parallel to each other. The earth's field of gravitational force 
is uniform for terrestrial distances, if we exclude from considera- 
tion the effects of the compression at the poles and the effects 
of its rotation. 

Example, Compare the attraction of the earth for the same 
mass at the height of i and of 100 cm. from its surface. 

Solution. The field of force within such small distances is 
uniform; the attraction is therefore the same. The weight will 
be the same at both points. 



FIELDS OF FORCE 1 59 

Radiant Fields of Force. A field of force established by a 
point such as a magnet pole of very small size varies in strength 
in accordance with the laws of radiant or central forces. If two 
points such as minute magnet poles act upon each other the 
action will vary inversely with the square of the distance sepa- 
rating them and with the product of the strength of the poles. 
The force exerted upon each other by two points such as magnet 
poles is expressed by the following formula 



in which i^ is the force, A/' is a constant, w and m' are the quanti- 
ties acting on each other, and / is the distance separating them. 
In the case of magnet poles N is unity; in the case of the 
attraction of gra\'itation between two masses of small size N has 
a value of 6.6576 X 10-^. In all cases where units of the C.G.S. 
system are used F will be determined in dynes. Thus a mass of 
I gram is attracted by another mass of i gram at a distance of 

I cm. by a force of 6.6576 X 10-® dyne. 

Example. A mass of 7 grams and a mass of 3 grams are 

II cm. apart. What is the amount of gravitational force they 
exert upon each other ? 

Solution. Owing to the attraction of gravitation they will 
attract each other. The amount of the attraction in grams is 
given by the formula, in which the values of the problem are 
substituted for the letters of the expression : 

F = 6.6576 X io-« X 'j^ = 6.6576 X 10-^ X 0.173 
= 1. 15 X I0-* dyne. 
Reciprocal Action in Fields of Force. The numerator of 
the fraction N -—^ expressing the value of the force con- 
sists of the product of two quantities and a constant. If either 
quantity m or m! is reduced to zero, the value of the fraction 



l60 ELEMENTARY ELECTRICAL CALCULATIONS 

will also become zero and no force will be exercised. All such 
force is exerted reciprocally. The earth attracts all things, but 
all things attract the earth. A magnet not only attracts its 
armature, but its armature attracts it. 

In the case of an electric field tn and mf are quantities of 
electricity or of magnetism, such as magnet poles. 

Example. Two equal magnet poles act upon each other with 
a force of 3.4 dynes. The distance between them is 2.5 cm. 
What is the strength of each ? 

Solution. By the conditions of the problem, m = m\ and the 

formula becomes F = — • Substituting for F and / their 

L 

values we have 

3-4 = 7 — Ti' or ni^ = (2.5)2 X 34 = 21.25 and m = 4.6. 
(2.5) 

Each pole therefore is equal to 4.6 unit magnet poles. 

Example. Two equally charged disks repel each other with a 
force of 10.4 dynes, the distance between them being 3.1 cm. 
Calculate the amount of charge on each, assuming them to be so 
small in proportion to the distance separating them that the laws 
of radiant action apply. 

Solution. Substituting in the formula as before gives 



10.4 = - — rr , or m^ = (3.1)2 y^ jQ^ ^ 99.944 and m = 10. 
(3-i>) 

Each disk is charged with 10 electrostatic units of electric 
quantity. 

Example. One of the preceding disks is placed 3.4 cm. from 
another similarly charged disk. It attracts it with a force of 2.1 
d)nies. What is the quantity of electricity upon the second disk? 

Solution. This is a case for the formula F= —zr- We have 

P 

m = 10, F = 2.1, and / =3.4. Substituting as before we have 
2.1= - — -, 10 m' = (3.4)2 y^ 2.1 = 24.276, and m' = 2.43. 



FIELDS OF FORCE l6l 

The new disk is charged with 2.43 electrostatic units of 
quantity. 

Example. A magnet pole attracts another magnet pole at a 
distance of i cm. with a force of 3 dynes. The one pole is a unit 
pole. What is the strength of field at unit distance from the 
other pole ? 

Solution. Here unit quantity is attracted with a force of 
3 units. The field at the unit pole, therefore, has a strength of 
3 units, or 3 lines of force per square centimeter. 

Induction of E.M.F. and Current. If a unit length of con- 
ductor is moved at unit rate across a unit field, a unit potential 
difference or e.m.f. will be maintained between its ends. The 
conductor is assumed to move at right angles to the polarity of 
the field. 

Example. An electro-magnetic field of force has 21 lines to 
the square centimeter. A wire 121 cm. long is moved across 
it at the rate of 32 cm. per second. What e.m.f. will be 
generated? 

Solution. The wire is 121 units long, and moves at 32 units 
rate; in a unit field it would have 121 X 32 = 3,872 units of 
e.m.f. impressed upon it. But the field is of 21 units intensity; 
therefore the e.m.f. impressed would be 3,872 X21 =81,312 
C.G.S. units, which are equal to 0.000,813 volt. 

Current actuated by e.m.f. thus produced is induced current. 
For a current to be produced there must be a closed circuit. If 
the quantity of field included within a circuit of wire or other 
conductor is changed, a current will be produced in the con- 
ductor. The current is due to e.m.f. impressed. The current 
and e.m.f. will both vary with the rate of change of quantity of 
field. A unit rate of change will impress unit e.m.f. on the 
circuit. The unit of rate of change is the product of a cross- 
sectional area of i cm. by an intensity of i line of force per square 
centimeter per second. 



1 62 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. A circle of wire is i meter in diameter. Within 
it there is a field of 257 lines of force to the square centimeter. In 
9 seconds it is reduced to 61 lines. What e.m.f. will be impressed 
on the circuit ? 

Solution. The area of the circle is 7,854 square cm. This 
multiplied by the rate of change of strength of field gives the 
e.m.f. The rate of change is the number of lines of force per 
square centimeter destroyed per second, or (257 — 61) -^ 9 = 
21.78. 7,854 X 21.78 = 171,060 C.G.S. units =0.001,710,6 volt. 

I E.M. unit of e.m.f. is produced by the cutting of i line of 
force per second, or by a change in the number of lines of force 
within a circuit at the rate of i line of force per second. 

I volt e.m.f. is produced by the same process, but with 10^ 
lines substituted for i line in the law. 

Example. A field of force has a strength of 39 lines per 
square centimeter. A conductor 1 1 cm. long moves across it at 
the rate of 17 cm. per second. What e.m.f. will be produced ? 

Solution. The lines of force cut per second by the conductor 
are equal to the area it sweeps through in a second multiplied by 
the number of lines in a square centimeter, which gives 

e.m.f. =11 X17 X39 = 7,293 C.G.S. electro-magnetic units 
= 0.000,072,93 volt. 

Example. A car goes at the rate of 30 miles an hour on a 
railroad of 4 feet 8 inches gauge. The vertical component of 
the earth's magnetic field may be taken as 0.438 line of force. 
What e.m.f. will be developed in an axle ? 

Solution. A speed of 30 miles an hour is 1,341 cm. per second. 
The length of the car axle, 4 feet 8 inches for the purposes of 
the problem, is 142.24 cm. Multiplying these together gives the 
area through which the axle sweeps in a second. It is 1,341 X 
142.24 = 190,743 square cm. Each square centimeter has its field 
of 0.438 line of force. Therefore 190,743 X 0.438 = 83,545 



1 



FIELDS OF FORCE 163 

is the number of lines of force cut per second and also the 
number of C.G.S. units of e.m.f. As a volt is equal to 10^ 

C.G.S. units, this is equal to 83,545 -^ 10^ = volt,approx- 

1200 

imately. 

A dynamo's or motor's working field is the portion swept 
through by the wires or conductors on the armature. In the 
case of a two-pole machine with a drum armature the con- 
ductors pass through the field twice in each revolution, but as 
they are connected so as to work in parallel half in parallel with 
half, the e.m.f. is due to the cutting of the lines of force of the 
field once for each revolution. 

A convenient formula for the e.m.f. of such machines is the 

foUowing: ^^^ 

e.m.f. m volts = — r-' 

in which n is the revolutions of the armature per second in a 
two-pole machine, or equivalent thereto in a multipolar machine, 
5 is the number of conductors in series on the armature, and N 
is the number of lines of force in the field cut by the armature 
conductors in a revolution. 

In a multipolar machine as usually wound for direct current 
the above value is to be multiplied by one half the number of 
poles if N is taken as the field due to one pair of poles. This 
is on the basis that a multipolar machine is usually an aggre- 
gation of bipolar ones. For such a multipolar machine the 

formula becomes tnSN 

e.m.f. in volts = ^- — — ' 

in which p is the number of pairs of poles. 

Example. From the pole pieces of a two-pole dynamo 
7,170,000 fines of force run to the armature; there are 120 turns 
of wire on the armature, all in series; the machine turns at the 
rate of 780 revolutions per minute. Calculate the e.m.f. 



l64 ELEMENTARY ELECTRICAL CALCULATIONS 

Solution. 780 revolutions per minute are 13 per second. 
Substituting in the formula gives 

c • u 1 3 X 120 X 717 X 10^ ,, 

e.m.f. m volts = -^ ~—^ = iii.g volts. 

Example. A dynamo has eight poles; there are 9,000,000 
lines of force in each of the fields ; the armature rotates at a rate 
of 20 revolutions per second; there are 90 conductors on the 
armature, all in series. Calculate the e.m.f. 

Solution. Substituting in formula (2) we have 

f ■ u 4 X 9 X 10^ X 20 X 90 A o u 
e.m.f. m volts = ^'^ ^— 2- = 648 volts. 

In such examples as these the use of exponential notation 
simplifies the work. Thus on inspection the first of the above 
examples reduces to 13 X 12 X 0.717 and the second to 4 X 9 

X 2 X 9- 

Multipolar machines are sometimes so constructed that 
the calculation follows the rule for bipolar machines, the 
lines of force in one of the fields being taken as the value 

of N. 

PROBLEMS. 

A field of 363 square cm. area contains 76,321 lines of force. What 
is its strength in gausses ? Ans. 210 gausses. 

A field of force has a strength of 1,100 lines to the square centimeter; 
its cross-sectional area is 36 square cm. How many lines of force does 
it contain ? Ans. 39,600. 

A field of force is specified as of 2,700 gausses and 297,000 webers. 
What is its cross-sectional area? Ans. no square cm. 

What is the strength of field in square-inch measure if it has 4,000 
lines of force to the square centimeter? 

Ans. 25,800 lines of force to the square inch. 

A field of force has 2,000 lines of force and is of 5 square cm. area. 
What will its action on a magnet pole be? 

Ans. 400 dynes; 0.408 gram; 6.30 grains. 



FIELDS OF FORCE 165 

An electro-magnet attracts a magnet pole of 29 units strength with a 
force of 5 grams. What is its strength of field ? 

Ans. 169 lines of force to the square centimeter; 1,090 lines of 
force to the square inch. 

Two equal magnet poles attract each other with a force of 0.005 
gram; they are 0.5 cm. apart. What is the strength of each ? 

Ans. 1. 107 unit poles each. 

Two disks attract each other with a force of 0.003 E^^^} ^^ch carry- 
ing the same static charge; they are 0.7 cm. apart. How many E.S. 
units of quantity does each one carry? Ans. 1.208 units. 

A magnet pole of 6 units strength attracts another one with a force 
of 0.0414 gram at a distance of 2 cm. What is the strength of the other 
magnet? Ans. 27.07 units. 

A magnet of 7.0 unit poles is attracted at a distance of 1.5 cm. by 
another magnet with a force of 0.31 gram. What is the field at a 
distance of i cm. from the other magnet ? 

Ans. 97.75 lines of force to the square centimeter; 630.5 lines of 
force to the square inch. 

A two-pole dynamo has 9,000,000 lines of force in its field; there are 
140 conductors in its armature. At 1,200 revolutions per minute what 
is its voltage ? Ans. 252 volts. 

The area of the pole face of a two-pole dynamo is 36 square inches; 
it is excited to 65,000 lines of force to the square inch; the armature 
has 200 turns of wire and revolves 1,500 times per minute. Calculate 
the e.m.f. Ans. 117 volts. 

There are 120,000 lines of force to the square inch in the field of an 
eight-pole dynamo; each pole face is 48 square inches area; on the 
armature there are 2,400 conductors in 8 parallel groups; the revolu- 
tions are 10 per second. Calculate the e.m.f. Ans. 691.2 volts. 

An e.m.f. of 200 volts is to be obtained from a two-pole machine 
whose pole faces are of 9 square inches area each; the induction is 
100,000 to the square inch; the rotation is to be 25 per second. How 
many conductors should there be on the armature ? 

Ans. 889 conductors (by computation). 



CHAPTER XIII. 



MAGNETISM. 

Moments. — Lever Arm of a Force. — The Couple. — Unit of Moment. 

— The Magnetic Filament. — Lines of Force in a Filament. — 
Magnetic Quantity and Strength of Pole. — Measure of Magnetic 
Quantity. — Lines of Force produced by Unit Quantity of Magnetism. 

— Moment of a Magnet. — Intensity of Magnetization. — Two Defi- 
nitions of Intensity of Magnetization. — Turning Moment of a Magnet. 

— Magnetic Traction. — Problems. 

Moments. If a bar or lever is pivoted at a point so as to turn 
with the point for a center, and if a force is applied to the bar at 
a distance from the center and at an angle to the bar so as to 
tend to turn it, the turning action of the force is termed its 
moment. 

The moment of a force is the product of the force by the 
length of the line connecting the center of rotation with the line 
passing through the point of application of and in the direction 
of the force and perpendicular thereto. If necessary, the line 
passing through the point of application of the force is prolonged. 
In the cut let A be the point around which as a center the lever 
^ C is free to rotate ; let the line DF represent the force applied 
P at the point D. Drawing the line AB per- 

pendicular to the line of the direction of the 
force DF, the product of the force by the length 
of the Hne AB is the moment of the force. The 
next cut shows the same condition of things, 
except that the line of the force has to be pro- 





longed. Similar letters are used to indicate the same things. 
DF is the force applied to the lever, AC rotating around the 

i66 



MAGNETISM 1 67 

point Af and AB is the lever arm of the force, and the prod- 
uct of AB by DF is the moment of the force. 

Lever Arm of a Force. In the discussion of moments the 
perpendicular distance from the center of rotation to the Hne of 
the force or to its prolongation is called the lever arm of the force. 
In the cuts the hnes AB are the lever arms of the forces DF. 
The Couple. When two forces concur to produce rotation in 
the way described above for a single force 
they are termed a couple. Sometimes the term 
* couple ' is restricted to the two equal and 
opposite forces as shown in the cut, F and 
F\ whose lever arms are AB and AB\ Such 
restricted use of the term applies to the action Bj-- 
of the two magnet poles of the same 
/t\" magnet when the magnet is mounted 
so as to be free to turn about its center. 
Unit of Moment. Unit of moment 
is a unit of force acting with a lever 
arm of unit length. When a force has a moment of 
say 10 units it may itself be of i dyne value with a 
lever arm of 10 om., or of 10 dynes value with a lever 
arm of i cm., or any relation may exist which gives 
^/- the product 10. 

The Magnetic Filament. A magnetic filament is a 
theoretical conception. If a thin filament of iron were mag- 
netized under such conditions that no lines of force leaked out 
of its sides, and if it were so long that the poles would not 
act upon each other, it would be a magnetic filament. This 
condition is approximately present in the central filaments, 
which may be assumed to exist in and to constitute the central 
part of the iron or steel of a magnet. 

Lines of Force in a Filament. If an isolated magnet pole 
be assumed to be a possible thing, it would radiate lines of force 




1 68 ELEMENTARY ELECTRICAL CALCULATIONS 

symmetrically in all directions like the radii of a sphere. The 
poles of a magnetic filament are assumed to do this and to that 
extent to represent isolated magnet poles as shown in the cut. 

Assuming the lines of force to radiate symmetrically in all 
directions from the magnet pole, the number of lines of force 
emerging from the pole are equal to the number of lines per 
square centimeter at the unit distance from the pole multiplied 
by the surface of the sphere of unit radius. From geometry we 
know that the surface of a sphere of radius i is equal to 4 7t. 
This is because the area of a sphere of radius r is equal to 4 Ttr^. 
But the square of i is i, and if we substitute for r the radius of 
the sphere in question we shall have 4 ^ri^, which is equal to 4 n. 
The strength of the pole is measured by the lines of force per 
square centimeter at unit distance. Therefore the lines of force 
emerging from the pole are equal in number to the product of the 
strength of the pole by 4 tt, which is the product of 4 X 3.1416 = 
12.5664. 

An indefinitely thin filament magnetized ever3rwhere in the 
direction of its length is called a magnetic solenoid. It follows 
that the total number of lines of force in a unit magnetic 
filament are 12.5664. 

Example. A pole of a magnetic filament establishes a field of 
10 Unes of force to the square centimeter at a distance of i cm. 
from itself. What is the number of fines of force emerging from 
the pole ? 

Solution. Multiplying the lines of force per square centimeter 
at unit distance by 4 n, we have 10 X 12.5664 = 125.64, the lines 
of force emerging from each pole after threading the filament. 

Example. Assume that a field of 9 lines of force per square 
centimeter is established by such a pole at a distance of 3 cm. 
Calculate the lines of force in the filament. 

Solution. Multiplying the lines of force per square centimeter 
at 3 cm. by the square of the distance gives the lines of force 



MAGNETISM 169 

per square centimeter at unit distance, or 9 X (sY — 81. 
Then multiplying the lines of force per square centimeter at 
unit distance by 4 tt gives the answer. 81 X 12.5664 = 
1,017.88. 

Magnetic Quantity and Strength of Pole. Magnetic 
quantity is the measure of the strength of a magnet pole. The 
strength of a magnet pole is its attractive or repulsive force for 
another pole of unit strength. A pole of unit strength is one 
which will attract or repel another pole of unit strength at a dis- 
tance of I cm. with a force of i gram. A unit pole contains 
I unit of magnetism. 

A magnet pole is a center of force whence lines of force issue, 
passing out from it into space. A magnet pole always has a 
complementary pole; for anorth pole there is always a south pole, 
and vice versa. The lines of force emerging from one pole 
always return to the other pole through space. Thus each line 
of force follows a continuous path or circuit. 

Quantity of magnetism may be defined as the magnetization of 
a magnet irrespective of its length, but the term ' magnetization ' 
is usually applied to the measure of the relative magnetic moment 
of a magnet, in which case the volume of the magnet enters as a 
constituent. Magnetic quantity is taken as located at the end 
or pole of a magnet. The unit of quantity of magnetism is the 
quantity which would act upon a unit pole at a distance of i cm. 
with a force of i dyne, so that the quantity of magnetism in a 
unit magnetic pole is the unit quantity of magnetism. Quantity 
of magnetism constitutes the strength of a magnetic pole, and 
the two expressions can often be used interchangeably. 

Measure of Magnetic Quantity. Magnetic quantity is 
measured by the strength of field which it can produce at a dis- 
tance of I cm. from its place or location, which is a magnet pole. 
The strength of field is measured by the dynes of force which it 
can exert upon a unit magnet pole. Hence the dynes of force 



I70 ELEMENTARY ELECTRICAL CALCULATIONS 

with which a unit pole is acted on by a quantity of magnetism 
I cm. distant give the units of magnetic quantity. 
Lines of Force produced by Unit Quantity of Magnetism. 

Recurring to the definition of a Hne of force we see that the 
unit of magnetic quantity produces a field of one Hne of force per 
centimeter at a distance of i cm. from the pole where it is located. 

Example. A magnet produces a field of 3 lines of force per 
centimeter at J cm. distance from its pole. What is the quan- 
tity of magnetism in a pole ? 

Solution. At I cm. from the pole it would produce a field of 
} Hne of force per centimeter. The quantity of magnetism is 
} of a unit. 

The reduction factor 4 used in the last problem is the 
square of the ratio of J to i ; this ratio is 2, and its square is 4. 

Example. A unit pole is placed at a distance of 3 cm. from 
a pole of unknown quantity of magnetism. The attraction 
between them is 9 dynes. What is the quantity of magnetism 
in the pole of unknown strength ? 

Solution. In the formula for radiant force, force = -— -^ 

(distance)'' 

substitute the values of the problem. This gives 

I X w' 
9 = » whence we find m'= 81, which is the quantity 

of magnetism in the pole. The pole of unknown strength has 
the strength of 81 unit poles. If it were placed at a distance of 
I cm. from a unit pole the attraction between them would be 
81 dynes. 

Example. Two magnets act upon each other with a force of 
3.6 d3nie3 at a distance of i.i cm. One of the poles acts upon a 
unit pole with a force of 2 dynes at a distance of 2.1 cm. What 
is the strength of the poles? 

Solution. The general formula can be transposed from 

Force = 



(distance)^ 



MAGNETISM 171 

in which m and m' are the quantities acting on each other, to 
^ force X (distance)^ 

Substituting for force 2, for distance 2.1, and for m' i, it 
becomes the case of the one pole acting upon the unit pole, 
giving 

which is the strength of one of the poles. Again applying the 
formula with this as the value of w' we have 

3-6 X (i.i)^ 
^== 8.82 ==^-^94, 

which is the strength of the other pole. 

Example. What strength of field would be established by 
a 1.75 pole at a distance of 0.33 cm.? 

Solution. The strength varying inversely as the square of the 
distance, the proportion obtains: 

(0.33)2 : I : : 1.75 : x = 16.07. 

At I cm. distance from the pole there are 1.75 Hnes of force to the 
square centimeter. At 0.33 cm. distance there are 16.07 lines. 
At the lesser distance a unit magnet pole would be acted on by 
16.07 dynes. 

Moment of a Magnet. The moment of a magnet is the 
product of the strength of one of its poles by the distance separat- 
ing the two poles, which is the same as the sum of the products 
of the strength of each pole by its distance from the center of the 
distance separating the poles. 

Example. The poles of a magnet are of 12.9 dynes strength 
each; they are separated by 14 cm. from each other. Calculate 
the moment of the magnet. 

Solution. It is 12.9 X 14 = 180.6. 



1/2 ELEMENTARY ELECTRICAL CALCULATIONS 

Another definition of the moment of a magnet is a quantity 
which when multiplied by the intensity of a uniform field gives 
the couple which the magnet experiences when held with its 
axis perpendicular to the lines of force in this field. 

Example. A magnet is pivoted so as to be able to turn with 
freedom. It is held in a field of force of 30 lines to the square 
centimeter, and at right angles to the lines of force. It tends 
to turn with a moment of 21. What is its moment ? 

Solution. It is 0.7, because 30 X 0.7 = 21. 

Example. If it had a strength of pole of 0.25 dyne, how far 
apart would the poles be ? 

Solution. Its moment being 0.7, its length from pole to pole 
is 0.7 -7- 0.25 = 2.8 cm. 

The turning effort, couple, or moment of a magnet at right 
angles to parallel lines of force of a unit field is the same as if 
each pole were acted on by a unit magnet pole at unit distance 
or at one centimeter and placed in a line perpendicular to the 
magnet. The action of a field of n lines is equal to that of a pole 
of quantity of magnetism n at a distance of i cm., or to that 
of a pole of quantity na^ at a distance a. 

Intensity of Magnetization. If a magnet were cut into 
pieces, the quantity of magnetism in the pieces would be the 
same as in the original magnet. The sum of the moments of 
the pieces would also be the same as the moment of the original 
magnet. 

The intensity of magnetization is the magnetic moment of a 
unit volume of the magnet under consideration. If a magnet 
however large were cut into pieces of unit volume each, the sum 
of the magnetic moments of the pieces would be equal to the 
magnetic moment of the original magnet. Intensity of magnet- 
ization is the quotient of moment divided by volume. It expresses 
the relative strength of a magnet. It is sometimes called magnet- 
ization. 



1 




MAGNETISM 173 

Example. A steel bar magnet weighs 453.6 grams. Its 
specific gravity is 7.85. Its moment is 10,088 C.G.S. units. 
What is its intensity of magnetization? 

Solution. The weight in grams divided by its specific gravity 
gives the volume in cubic centimeters. Hence 453.6 -^ 7.85 = 
57.78 is the volume of the bar. The quotient of the moment 
divided by the volume gives the intensity of magnetization, or 
10,088 -r- 57.78 = 174.6, which is the intensity of magnetization. 

Intensity of magnetization is also equal to the quantity 
of magnetism per square centimeter of cross section of the 
magnetized piece. Thus let m denote the quantity of magnetism 
upon a pole of cross section 5, and let / denote the distance from 
pole to pole. The moment of the magnet is then ml. Its 
volume is SI. Dividing moment by volume to obtain the moment 
per unit volume, which as we have seen is the intensity of mag- 
netization, we have 

Intensity of magnetization = — - = — ■» 

which is the magnetic quantity per unit area of cross section of 
the bar. 

For the above to be true the magnet must be magnetized 
everywhere in the direction of its length. Such distribution of 
magnetism is termed solenoidal. 

Example. A bar magnet is 1.2 cm. in diameter, 16 cm. long, 
and is of 90 dynes polar strength at a distance of i cm. Calculate 
the intensity of magnetization (a) by the magnetic quantity per 
unit area of cross section and (b) by the moment of unit volume. 

Solution. The cross-sectional area of the bar is 0.6^ X 3.1416 
= 1.13 square cm. The intensity of magnetization is the quotient 
of the quantity of magnetism, 90, divided by the area of the 

surface over which it is distributed, 1.13, or -^ — = 79.64. 

o 
The moment of the bar is the product of the length, 16, by 



1/4 ELEMENTARY ELECTRICAL CALCULATIONS 

the quantity of magnetism on one pole, 90, giving 16 X 90 == 
1,440. The volume is the product of the area, 1.13, by the 
length, 16, giving 1.13 X 16 = 18.08 cubic cm. Dividing moment 
by volume we have 

^^^ = 79.64, as before. 

Two Definitions of Intensity of Magnetization. Intensity 
of magnetization is thus definable from two standpoints, that 
of magnetic quantity and that of magnetic moment. Referred 
to magnetic quantity it is equal to the quantity in a magnet pole 
divided by the area of the face of the pole. It is therefore 
equal to the number of units of quantity or of unit poles per 
square centimeter of the area of the pole. This area is the end 
face of the magnet at right angles to the axis. Referred to 
moment it is equal to the moment of a cubic centimeter of the 
magnet. As we have seen, one of these expressions is equal to 
the other. 

Turning Moment of a Magnet. If a magnet is turned so 
as to be at an angle with the Hnes of force of a field, it will tend 
to place itself parallel to the Hnes of force. The couple is equal 
to the sum of the turning moment of the two poles. As the poles 
are of equal strength and are generally symmetrical, the couple 
is numerically equal to twice the turning moment of one pole. 
Referring to what has been said about couples and moments 
(pp. 166, 167) it will be seen that the couple of a magnet in a 
field of force is numerically equal to the product of the strength 
of one pole by the length of the magnet by the sine of the angle 
between the axis of the magnet and the lines of force. 

As the sine of 90° is i, the couple of a magnet at right angles 
to the lines of force of a field is equal to the product of the strength 
of one pole by the length of the magnet. In other words, the 
couple of a magnet at right angles to a field of force, which 



MAGNETISM 1 75 

means at right angles to the lines of force of such field, is equal to 
the moment of the magnet. 

Example. A magnet held at right angles to a unit field of 
force has a couple of 22. What is the strength of one pole in 
dynes if it is 25 cm. long ? 

Solution. The magnet's length being 25 cm., the moment is 
25 times as great as if the magnet were i cm. long. The strength 
of one pole is therefore |f = 0.88. To prove the correctness 
of the method refer to the law that the moment of a magnet is 
equal to the product of the strength of one pole by the length of 
the magnet. The above operation reverses the steps indicated 
by this law. 

Example. If the above magnet were placed as described in 
a field of force of a strength of 3.7, what would the couple be ? 

Solution. It would be the product of the pole strength by the 
length of the magnet by the strength of the field. 0.88 X 25 X 
3.7=8.14. 

Example. A magnet is placed in a field of force at an angle 
to the lines of the field of 45°. The magnet is 2.8 cm. long. 
What is the lever arm of its couple ? 

Solution. Accurately speaking there are two arms. Each 
is equal to the length of one half of the magnet multiplied by the 
sine of the angle which it makes with the lines of the field. The 
sine of 45° is 0.7071 and the length of one arm of the magnet is 
2.8 -T- 2 = 1.4. 1.4 X 0.7071 = 0.99, which is the lever arm 
of each pole. The couple is equal to the sum of the turning 
moments of the two poles. The turning moment of a pole is the 
product of its strength into the lever arm of its action, in this 
case 0.99. 

Magnetic Traction. The formula for the traction between a 
magnet and its armature when the two are in contact is deduced 
on pages 192-194. It is 

Traction = — — > 

St: 



1/6 ELEMENTARY ELECTRICAL CALCULATIONS 

in which A is the area of the face in contact with the armature 
and B is the lines of force that pass through one square centimeter 
of the area of contact. 

Example. A magnet with a face area of i.i square cm. and 
250 lines of force to the square centimeter is in contact with 
an armature. Calculate the traction. 

Solution. Applying the formula we have 

Traction = -7— — ^ ■ = 2,736 dynes = 2.789 grams. 
8 X 3.1416 '^ •' ' 7 o 

PROBLEMS. 

What is the number of lines of force in a magnetic filament which 
produces a field of force of 11 lines of force to the square cm. at a 
distance of ^.^ cm. from one of its poles ? Ans. 1,505 lines of force. 

A filament has in it 90 lines of force. What field will it produce at 
I cm. from one of its poles? 

Ans. 7.162 lines of force to the square centimeter. 

What quantity of magnetism is in the magnet of the last problem ? 

Ans. 7.162 units. 

If a magnet had 19 units of magnetism, what field would it produce 
at a distance of 19 cm. from one of its poles? 

Ans. 0.526 lines of force to the square centimeter. 

What is the magnetic quantity of a magnet pole if the magnet attracts 
a 17-unit pole at a distance of 5 cm. with a force of 3 dynes? 

Ans. 4.41 units. 

Two poles of unknown strength act on each other with 1.3 milligrams 
force at a distance of 4 cm. One (a) acts on a unit pole at a distance of 
3 cm. with a force of 2 milligrams. Calculate the strength of the two 
poles. Ans. (a) 17.66 units; (b) 1.15 units. 

Calculate the moment of a lo-inch magnet with 10 milligrams 
strength of pole. Ans. 249.2 C.G.S. units. 

A magnet in a field of force of 91 lines of force to the square centi- 
meter at right angles to the lines tends to turn with a force of 
82 dynes at each pole; it is 25.4 cm. long. What is its moment? 

Ans. 22.9 C.G.S. units. 



MAGNETISM 1 77 

A magnet is 0.3 square cm. in cross-sectional area, is 20 cm. long, 
and at i cm. distance attracts a unit pole with a force of 3 grams. 
Calculate its intensity of magnetization by two methods. 

Ans. (a) 2,943 dynes -^ 0.3 = 9,812 C.G.S. units, (b) Moment 
= 58,860; volume = 6 cubic cm.; 58,860 4- 6 = 9,810 C.G.S. units. 

A magnet is 12 cm. long; it is placed in a field of force of 11 lines to 
the square centimeter at an angle of 30° to the lines of force; it has 
30 units of magnetism. What is the couple ? 

Ans. 1,980 C.G.S. units. 

If a field of 300 lines of force to the square centimeter were to be 
replaced by a magnet of 29 units strength, at what distance from the 
pole would the original field exist? Ans. 0.31 cm. 

How far from a point would a lo-unit magnet pole have to be to rep- 
resent in its action at the point a 1.3 field of force ? Ans. 2.8 cm. 

What is the intensity of magnetization of a magnet weighing 395 
grams, with a moment of 11,721, taking its specific gravity as 7.85? 

Ans. 232.9 units. 

There are 1,000 lines of force to the square centimeter in a magnet; 
its pole area is 1.3 square cm. When in contact what will the traction 
be between it and its armature? Ans. 51,725 dynes; 52.7 grams. 

The area of contact between a permanent magnet face and armature 
surface is 2.9 square inches; 15,000 lines of force to the square inch 
pass through the area of contact. Calculate the traction in dynes, 
grams, and pounds. 

Ans. 4,025,130 dynes; 4,103 grams; 9.05 pounds. 



CHAPTER XIV. 
ELECTRO-MAGNETIC INDUCTION. 

Induction of Magnetism. — Relation of Induced Magnetization to Field. 

— Susceptibility. — Table of Susceptibility. — Magnetic Induction. 

— Permeability. — Permeance. — Reluctance. — Permeance and 
Reluctance. — Formulas for Inch Measurements. — The Magnetic 
Circuit. — Ampere Turns. — Intensity or Strength of Field Referred 
to C.G.S. Unit Turns. — Strength of Field Referred to Ampere 
Turns. — Total Field Referred to Ampere Turns. — Reluctance of 
Air. — Magneto-motive Force, — Intensity of Field at Center and 
Ends of Coil Interior. — Magnetic Circuit Calculations. — Reluctance 
of Circuit of Iron. — Ampere Turns for a Given Field. — Magnetic 
Traction. — Determination of Permeability from Traction. — 
Problems. 

Induction of Magnetism. Assume a field of force to be 
produced in the air or in a vacuum, and let a piece of iron be 
placed in the field. The original field will exist in it exactly as 
if it were air or a vacuum, and in addition thereto magnetism will 
be induced in the iron, so that more magnetism will be present 
in a unit cross-sectional area of the iron than in a corresponding 
area of the field. The additional quantity of magnetism per 
unit cross-sectional area is termed intensity of induced magnet- 
ization, or simply induced magnetization and is indicated by I. 
The magnetic intensity of the inducing field per unit of cross- 
sectional area being indicated by H, the total magnetism of the 
iron is equal to its cross-sectional area multiplied by the total 
quantity of magnetism per unit area, and as the latter is the sum 
of JT + J, if we call the area A the total magnetism in the iron 
will be ^ X {B:-\- I), 

Intensity of induced magnetization is measured by units of 
magnetic quantity, one of which units is the quantity in a unit 
magnet pole. 

Relation of Induced Magnetization to Field. — Suscepti- 
bility. The value of induced magnetization depends on the 

.178 



ELECTRO-MAGNETIC INDUCTION 1 79 

value of the magnetizing force which induces it, exactly as the 
current due to e.m.f . depends on the value of the e.m.f., with one 
difference. The relation of magnetizing force to induced mag- 
netization is expressed by the quotient of induced magnetization 
divided by magnetizing force. The quotient is called suscep- 
tibility and is indicated by ic (the Greek letter kappa) or by the 
letter K. The relation is expressed as far as susceptibility is 
concerned by the two expressions 

K =-~ and I = kH. 
H 

If K were conductivity, -H" electro-motive force, and I current, 
the above formulas would correspond in form to Ohm's law. The 
analogy fails and the difference just spoken of appears because 
of the law that the value of k changes for different values of I. 

The value of /c is also different for different irons, and has to 
be determined experimentally for each. 

Example. A bar of iron is exposed to a magnetizing force of 
1.7, and each element of i square cm. cross-sectional area has 
induced magnetization of 49.9 units induced in it. What is the 
susceptibility of the iron at the given excitation (49.9) ? 

Solution. The magnetizing force per square centimeter of 
field is 1.7, which is jff of the formula; the induced magnetization, 
49.9, is I. Substituting in the formula we have 

I 49.9 

which is the susceptibility of the iron at the given excitation. 

Example. In a field of -HT = 3.5 a pole strength of 172.2 
dynes is induced in a bar of 0.3 square cm. area. What is the 
susceptibility of the iron of which the bar is composed ? 

Solution. The pole strength divided by the area of the bar 
is the intensity of magnetization. 172.2 -5-0.3 =574 = I* 



l8o ELEMENTARY ELECTRICAL CALCULATIONS 

I divided by -H" gives the susceptibility, 574 -4- 3.5 = 164, which 
is the value of /c, or the susceptibility. 

Table of Susceptibility. The following table gives the 
values of susceptibility for different values of induced magnetiza- 
tion in wrought iron, according to experiments by Ewing and 
Bidwell. 



I 


K 


I 


K 


3 


10 


1,173 


"1 


32 


23 


1,249 


56 


117 


53 


1,337 


17 


574 


164 


1,452 


7 


1,078 


187 


1,530 


2.6 


161 


1,660 


0.067 



Example. What field is required to induce a magnetization of 
917? 



Solution. The formula H 



with the values of the prob- 



lem and of the table substituted becomes 

which is the magnetizing force. 

Susceptibility is also called the coefiScient of induced mag- 
netization. 

Magnetic Induction. The magnetic intensity of the field, 
as we have seen, is generally expressed in gausses or in lines of 
force per unit of cross-sectional area of the field, one line of force 
per square centimeter being the expression for the unit strength 
of field. The strength of field due to a magnetic filament at i cm. 
distance from its pole is equal to as many lines of force per square 
centimeter as there are units of magnetism in the pole. The 
lines of force in the filament are equal to the intensity of the 
field at I cm. distance from the pole multiplied by 4 ;r (12.566). 



ELECTRO-MAGNETIC INDUCTION l8l 

If induced magnetization is multiplied by 12.566, the product 
will be the lines of force per square centimeter in the iron due to 
such induced magnetization. But there are also present in the 
iron the lines of force of the original field, designated by H] 
therefore the total number of lines of force per square centimeter 
in the iron is the sum of these two quantities. The sum is 
JBr+ 12.566 jr. 

This quantity is called magnetic induction and is designated 
by J5. 

Example. A field of intensity 12.5 acts upon an iron core of 
susceptibility 98. Calculate the lines of force per square centi- 
meter in the iron. 

Solution. The induced magnetization is equal to the product 
of the field by the susceptibility, or 12.5 X 98 = 1,225. This is 
the value of I. The lines of force per square centimeter in the 
iron are equal to the sum of the intensity of field and the product 
of the induced magnetization by 12.566, which gives 
12.5 + (12.566 X 1,225) = iS>4o6, 

which is the value of B and is expressed in lines of force per 
square centimeter of cross-sectional area of the core. 

Permeability. Returning to the formula for magnetic induc- 
tion and remembering that I = fcH, the formula can be written 
in two ways, thus 

H + 12.566 J, 

and substituting for I its value, fcIT, and introducing the sym- 
bol B, 

B = H+ 12.566 fcs, otH + 4 TTKH = HX (1 + 4 tt/c). 

The compound factor (i + 4 tt/c) is called permeability; its 

reciprocal, — , is called reluctivity. The symbol of per- 
meability is /x. In engineering calculations permeability is the 
foundation of most of the work affecting magnetic circuits; sus- 



1 82 ELEMENTARY ELECTRICAL CALCULATIONS 

ceptibility is the basis of the theory of permeability, but is less 
used in practical calculations. 

Example. A field of intensity B" = 5.1 acts upon a sample 
of iron of susceptibility k = 169. Calculate the permeability 
and induction. 

Solution. The permeability is i + 471/^= 1 + (12.566X169) 
= 2,125. The induction is the product of the permeability by 
the intensity of field, or 2,125 X 5.1 = 10,838, which are the 
lines of force per square centimeter of cross-sectional area of 
the iron. 

Example. What is the reluctivity of the above iron ? 

Solution. It is the reciprocal of the permeability; 

I I 

— = • = 0.00047. 

fi 2,125 

Permeability varies in different irons, and also varies as the 
lines of force per given cross-sectional area differ in number. 
The values of permeability for two samples of iron at different 
values of induction are given in the tables. 

Permeance. Permeance stands in the same relation to per- 
meability as that occupied by resistance with reference to 
resistivity or specific resistance. It is the power of a specified 
circuit or portion of a circuit for carrying lines of force. The 
substance carrying the lines of force is a geometrical solid. If 
prismatic or linear in shape the permeance will vary directly with 
the cross-sectional area and inversely with the length, being 

equal to 

_, .... ^cross-sec. area A 

Permeability X : -7 ' or // y > 

A and / indicating cross-sectional area and length respectively. 

Reluctance. Reluctance is the reciprocal of permeance, and 
is therefore expressed by 



ELECTRO-MAGNETIC INDUCTION 



183 



Tables of Permeability. The following tables give values 
of permeability for different values of magnetic field. 



SQUARE CENTIMETER MEASUREMENT. 



Annealed Wrought Iron. 


Gray Cast Iron. 


B 


/^ 


H 


B 


f^ 


H 


5,000 
9,000 
10,000 
I I , 000 
12,000 
13,000 
14,000 
15,000 
16,000 


3,000 

2,250 

2,000 

1,692 

1,412 

1,083 

823 

526 

320 

161 

90 

54 

30 


1.66 

4 

5 

6.5 

8.5 

12 

17 

28.5 

50 
105 
200 

350 
666 


4,000 
5,000 
6,000 
7,000 
8,000 
9,000 
10,000 
I I , 000 


800 
500 
279 
^33 
100 

71 
53 
37 


5 
10 

21-5 

42 

80 
127 
188 
292 


17,000 
18 000 














19,000 
20,000 























SQUARE INCH MEASUREMENT. 



Annealed Wrought Iron. 


Gray Cast Iron. 


B 


/^ 


H 


B 


/^ 


H 


30,000 
40,000 
50,000 
60,000 
70,000 
80,000 
90,000 
100, 000 


4,650 

3,877 

3,031 

2,159 

1,921 

1,409 

907 

408 

166 

76 

35 

27 


6.5 
10.3 
16.5 
27.8 

36.4 
56.8 
99.2 

245 

664 
1,581 
3,714 
5,185 


25,000 
30,000 
40,000 
50,000 
60,000 
70,000 


763 
756 
258 
114 

74 

40 


327 
39-7 

155 

439 

807 
1,480 








110, 000 








120, 000 








130,000 
140,000 





















1 84 ELEMENTARY ELECTRICAL CALCULATIONS 

In a magnetic circuit there are often included portions of 
different permeance. The reciprocal of the sum of the recipro- 
cals of the permeances of the different portions of the circuit is 
the total permeance. By using the property of reluctance in the 
calculations the use of reciprocals is avoided. The reluctance 
of a magnetic circuit is equal to the sum of and is obtained by 
adding together the reluctances of its parts. This operation 
takes the place of the one just described. 

Example. A bar of iron has a permeabiHty of 2,079. It is 
circular in section, i inch in diameter, and 16 inches long. What 
is its permeance ? 

Solution. The area of the bar is 0.7854 square inch; i square 
inch is equal to 6.45 square cm. 6.45 X 0.7854 = 5.066 square 
cm., the area in square centimeters. The length of the bar in 
centimeters is equal to 16 X 2.54 = 40.64 cm. Substituting in 
the formula we have 

I1-; = 2,079 X r = 259.15, which is the permeance. The 
/ 40.64 

reluctance is =0.00^86. 

259-15 
Permeance and Reluctance. Formulas for Inch Meas- 
urements. The formula for permeance is, for centimeter 

measurement, fx — - To reduce this to inch measurements is 
1/ 

to put it into such form that if the length of the core is given 
in inches, and if the area of the core is given in square inches, 
it will give the permeance directly. Taking the area of a 
square inch as 6.45 square cm., and the length of an inch as 2.54 
cm., the formula becomes, for inch measurements, 

6.45 U.A U.A 

^ ^, = 2.54 ^. 
2.54/ ^^ I 

Example. Calculate the permeance of a bar of iron 10 inches 
long and i square inch in area, whose permeability is 1,200 
under the assumed conditions. 



ELECTRO-MAGNETIC INDUCTION 185 

Solution. The quotient of the product of the cross-sectional 
area by the permeability divided by the length is (1,200 X i) 
-7- 10 = 120. This multiplied by the factor 2.54 gives 304.8, 
which is the permeance. 

Or substituting directly in the formula we have 

_. ^^ 1,200 X I o 

Permeance =2.54X ~ = 304.8. 

10 

The formula for reluctance is, for centimeter measurement, — - • 

To reduce this to inch measurement, / and A must be multiplied 
by the factors which will reduce linear inches and square inches 
to centimeters also linear and square, as before. Intro- 
ducing these factors into the reluctance formula it becomes 

—^ — -=0.304 — -• This formula can be used when the 
6.45 M M 

dimensions are given in inches. 

Example. The mean length of the core of a magnetic circuit 
is 18 inches; the core is circular in section and 4.3 square 
inches in cross-sectional area. The permeability, at the value 
of B" employed, is 2,200. Calculate the reluctance. 

Solution. The quotient of the length divided by the product 
of the cross-sectional area and permeability is 18 -f- (4.3 X 2,200) 
= 0.0019. This has to be multiplied by 0.394, because the 
dimensions are given in inches. 0.0019 X 0.394 = 0.000748, 
which is the reluctance of the magnetic circuit. 

Or substituting in the formula the values of / and A, it becomes 

18 

Reluctance = 0.394 = 0.000,748. 

2,200 X 4.3 

The Magnetic Circuit. The magnetic line of force always 
forms a closed curve, regular or irregular in shape, and a collec- 
tion of lines of force due to the same field are in a general way 
parallel to each other or concentric. They define a path which 
is called the magnetic circuit. It is in its laws analogous to the 



1 86 ELEMENTARY ELECTRICAL CALCULATIONS 

electric circuit. The lines of force may lie within a core of iron 
or may be partly within the metal and partly outside of it. 

The field is always produced by a coil of wire through which 
a current of electricity is passed. A mass of iron within the coil 
by its permeance increases the flow of lines of force and is called 
the core. This outHnes the arrangement practically in univer- 
sal use in the construction of generators, motors, and other 
machinery of that type. 

Ampere Turns. The product of the number of turns in a 
field coil by the amperes of current passing through its wire is 
called the ampere turns. The strength of field varies with the 
ampere turns. If the current is expressed in other units, other 
compound units result, such as C.G.S. unit turns. Such units 
are Uttle used. 

Example. A field coil is wound of wire, which with its insula- 
tion is 0.160 inches in thickness. In the operation of the 
machine it carries a current of 9.4 amperes. The internal 
diameter of the coil is 0.80 inches; the external diameter is 2.80 
inches; its length is 3 inches. What is the number of the 
ampere turns? 

Solution. Half the difference of the diameters gives the thick- 
ness of the coil; it is (2.80 — 0.80) -?- 2 = i inch. In an inch 
there are 6 turns of wire, as there can be no fractional turns. The 
length of the coil gives room for 18 turns, and the thickness for 
6 turns, a total of turns for the coil of 6 X 18 = 108 turns. As 
the wire carries a current of 9.4 amperes, the product of 108 X 9.4 
= 1,015.2 ampere turns is the answer. 

Intensity of Field Referred to C.G.S. Unit Turns. The 
intensity of field produced by a coil whose axial length is great 
compared to its diameter varies at different points. It is the 
field in the interior of the coil that produces the magnetic circuit. 
It is proved that the intensity of the field in the interior of the coil 
at a part equally distant from each end is equal to the product 



ELECTRO-MAGNETIC INDUCTION 1 87 



of the current in C.G.S. units by the turns in the coil and by 47: 
divided by the axial length of the coil. If we call the current 
strength in C.G.S. units i\ and call the number of turns in the 
coil Sj the above statement becomes in the shape of a formula, 

Intensity of field = — - — • 



^H All the above refers to a coil without a core of any magnetic 

^m substance. 

^B Example. What is the intensity of field in the center of a coil 

^B 115 cm. long, with 550 turns, when it is passing a current of 

H II C.G.S. units ? 

^B Solution. In this case S = 550, t = 11, and / = 115. We 

^B will assume the coil to be of relatively small diameter. Our 

formula becomes 

T* V £ £ ij 4 X 3-1416 X 550 X II .. 
Intensity of field = "^—^ — =661.1. 

These are the lines of force per square centimeter of cross- 
sectional area at the center of the coil, i.e., the dynes of force with 
which the field would act upon a unit magnet pole placed at a 
point on the axis of the coil equally distant from each end. 

Strength of Field Referred to Ampere Turns. If we 
denote the intensity of a current in amperes by i, and take 12.566 
as the value of 4 n, and remember that an ampere is equal to 10-^ 
C.G.S. unit, the formula becomes for amperes 

., rr^^ 12.^66X81 1.2^66X51 

Intensity of field = — "^ -^ lo = — ^-= 

Example. Assume the preceding case but expressing the 
current in amperes, and calculate the lines of force per square 
centimeter at the center of the coil. 

Solution. We have as before, S = 550, / = 115, and the 
current strength, iy is no amperes. This is because, as i C.G.S. 



1 88 ELEMENTARY ELECTRICAL CALCULATIONS 

unit is equal to lo amperes, ii C.G.S. units are equal to no 
amperes. Substituting in the formula gives 

Intensity of field ^ ^-^566X550X110 ^ ^^^^ 

115 
as before. 

Total Field Referred to Ampere Turns. The rules just 
given are for determining the relative strength of field. The 
total number of lines of force within the coil will obviously depend 
upon the cross-sectional area. If the intensity of field be multi- 
plied by the cross-sectional area, the result will be the total num- 
ber of lines of force. Instead of multiplying the expression by 
the symbol of the cross-sectional area, which may be A^ the 
denominator of the fraction may be divided by it. The result is 
identical. For the total lines of force at the central part of the 
interior of a magnetizing coil this gives 

Total lines of force = hl5^><A^ x A, or'-^^^^' ■ 

The expressions are given for amperes only. 

Reluctr^nce of Air. The denominator of the fractional 
expression last given is the reluctance of the column of air in the 
interior of the coil. This denominator is IjA. 

Magneto-motive Force. The numerator of such expressions, 
0.4 TzSi^ is sometimes called magneto-motive force. 

Intensity of Field at Center r.nd Ends of Coil Interior. 
The intensity of field at the ends of the coil is one-half as great as 
that at the center. 

Example. A magnetizing coil is 229 cm. long; it has 4,400 
turns, and a current of 22.5 amperes passes through it. The 
cross-sectional area of the core space of the coil is 11 square 
cm. Calculate the lines of force at the axial center. 

Solution. Applying the formula for intensity of field gives 

T^ -^ t e. yA 1-2566 X 4,400X22.5 
Intensity of field = — =543.2. 



ELECTRO-MAGNETIC INDUCTION 189 

Multiplying by the cross-sectional area gives 
543.2 X II =5,975 = total lines of force in the center of the 
field. 

The numerator of the above expression may be treated as 
magneto-motive force and the quotient of 1/ A as reluctance. 
Thus 

Magneto-motive force = 1.2566 X 4,400 X 22.5 = 124,403. 

Reluctance = 229 -f- 11 = 20.82. 

Dividing magneto-motive force by reluctance gives 

Total lines of force of field = ^r— ^ = 5,975. 

20.82 

Magnetic Circuit Calculations. If an iron core forming 
a complete magnetic circuit be placed within a magnetizing coil, 
and if its permeance be so high that the lines of force follow it 
without leakage, the magnetic flow will be uniform throughout 
the core. The formula given for the lines of the field within the 
coil at its center now applies to all parts of the core, with two 
differences — the mean length of the core is taken as the length / 
of the formula, and the area has to be multiplied by the per- 
meabihty : 

Lines of force or mametic flow = -^-7^ — \ 

l/liA 

Example. An iron ring is 24 cm. in internal diameter; it is 
made of a cylindricri bar of metal 3 cm. in diameter. It is 
wound with a coil of insulated wire of 127 turns. A current of 
3 amperes is passed through the wire. The permeability of the 
iron at the given value of Jffis 2,300. Calculate the lines of force 
in the ring under the supposition that there is no leakage. 

Solution. The magneto-motive force is 1.2566 X 127 X 3 
= 478.764. The reluctance is the quotient of the mean length 
of the ring by the product of its cross-sectional area by 2,300. 
The mean length is the product of the mean diameter, 24 -|- 3 



190 ELEMENTARY ELECTRICAL CALCULATIONS 

= 27, by n. This product is 27 X 3.14 = 84.78. The cross- 
sectional area is the product of the square of one-half the diameter 
of the bar by tt. 1.5 (half the diameter of the bar) squared =2.25 
and 2.25 X 3.14 = 7.07, the cross-sectional area of the bar. 
7.07 X 2,300 = 16,261. This i^ jxA. Dividing the length by 
this gives 84.78 -i- 16,261 = 0.0052, which is the reluctance of 
the ring. Finally, 478.764 -^ 0.00522 = 91,711, the lines of force 
which thread the iron passing around the ring. 
Reluctance of Circuit of Iron. If for the denominator of 

the expression ■ ' . — just given, which denominator is 1/ [iA, 

and which is the reluctance of the circuit, the expression 
0.394 1/ ixA be substituted, the formula will apply to inch meas- 
urements. It becomes 

Lines of force or magnetic flow = — ' - , , — 7= ■■ ^' / ^ • 

0.394 l/fiA l/iiA 

Example. The length of a core of an electro-magnet and of 
its armature forming a complete circuit is 33.6 inches. The 
cross-sectional area is i.i sq. in. At a permeancy of 2,300 what 
is the total number of lines of force when excited by 381 ampere 
turns? 

Solution. The ampere turns, 381, multiplied by 3.19 equal 
1,215.39. This is the dividend or numerator. The length of 
core is 33.6, which has to be divided by the product of the 
area, i.i, by the permeability, 2,300, w^hich product is 2,530. 
33.6 -^ 2,530 = 0.01328. This is the divisor or denominator of 
the expression. Then 1,215.39 -j- 0.01328 = 91,520, which is 
the magnetic flow. 

We may substitute the values given directly in the formula, 
obtaining as the value of the magnetic flow 

^.10 X 381 , , 

— ,7 — = 91,520, as before. 

ZZ^^ - (2,300 X i.i) ^ '^ 



ELECTRO-MAGNETIC INDUCTION 191 

Ampere Turns for a Given Field. The problem most 
called for in electric work where the magnetic circuit is con- 
cerned is the determination of the ampere turns required to 
produce a given field. If we caU the lines of force of the field, or 
the magnetic flow, -F, we have from the preceding section 

F = -^, — - — for centimeter measurements. 
l/liA 

In this equation Si are the ampere turns. By dividing and 

transposing, we obtain ^p 

Si = 



1.2566 /jlA 

The ampere turns are equal to the quotient of the length of 
the core multiplied by the fines of force divided by the product 
of the permeability by the cross-sectional area by 1.2566. 

Example. What is the number of ampere turns which would 
be required to produce 440,000 lines of force in a bar of iron 
25.4 cm. long and 25.8 sq. cm. cross-sectional area, at a per- 
meability of 166? 

Solution. The numerator of the fraction is the product of 
the length of the core by the number of lines of force, or 
25.4 X 440,000 = 11,176,000. The denomincitor is the product 
of the cross-sectional area by the permeability and by 1.2566, 
or 166 X 25.8 X 1.2566 = 5,381.76. Dividing we have 

11,176,000 -^ 5,381.76 = 2,076, 

which are the ampere turns required. 

IP 

If in the formula Si = -; — - > which is for centimeter 

1.2566 fxA 

measurements, / is multiplied by 2.54, which is the number of 

centimeters in an inch, and if ^ is multiplied by 6.45, which is 

the number of square centimeters in a square inch, it gives 

^ 1.2566/1X6.45^ °*^'^^ m' 



192 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. Calculate the ampere turns required to force 
440,000 lines of force through a core of iron 10 inches long, 
4 sq. in. in area, and of 166 permeability. 

Solution. The product of the length by the lines of force IF 
is 10 X 440,000 = 4,400,000. This is the numerator of the 
fraction. The product of the permeability by the cross-sectional 
area /z^l is 166 X 4 = 664. Dividing we have 

4,400,000 -^ 664 = 6,626.5, 

and multiplying by the factor 0.3133 we have 

6,626.5 X 0.3133 = 2,076 

which are the number of ampere turns required. 

Magnetic Traction. The traction or attractive force exerted 
between a magnet face and another magnet face or armature 
with which it is in contact is expressed by the formula 

Traction ^ -^^^:-n 

8 n 

which is thus deduced : 

A magnet face may be regarded as a collection of unit magnet 
poles each of which is attracted by a force of i dyne in a unit 
field and by a force of B dynes in a field of intensity B. I has 
been defined as the intensity of magnetization ; its numerical value 
may be taken as expressing the number of unit poles in a square 
centimeter of cross-sectional area of the magnet core at any given 
place. Thus the action of a unit field on a magnet face of i sq. 
cm. area is equal to I, and on a face of A square centimeter area 
is equal to Al. 

When a magnet and armature are in contact the field is made 
up of two component parts equal in value, each one due to one 
of the faces in contact. The action is therefore the product of 
the field due to one of the faces multiplied by the equivalent of 
the unit magnet poles represented by the other face. 



ELECTRO-MAGNETIC INDUCTION 193 

Let the strength of field be H, as usual ; then as we have seen 
the induction will h& H -{-^tzI = B, by the usual nomencla- 
ture. The portion of the induction due to the faces of the 
magnet and armature is ^nl = B — H, and that due to a 
single face is J (JB — H), It is obvious that this portion cannot 
act upon itself, so the attraction is the product of the magnetism 
in one face, Al^ acted on by the rest of the field intensity, 
which IS B - i {B - II) = ^ {B + H). From the equation 
H-\- 4T[I= B, we obtain a value of I in terms of B and BC, 

47? ^ 4;r 

i{B + M)= ^ii^-l^ ^ so that 
o rr 

Traction = — ^— • (i) 

8 rr 

In the above formula the attraction is given in dynes; as a 
dyne is approximately ^ix gram, the formula for grams is 

Iraction = ; srams, (2) 

24,655 ^ ' ' ^ 

in which the denominator is the product of 981 X 8 ;r. 

The area of contact being expressed in inches the area A of 
the formula has to be multiplied by 6.45, the number of square 
centimeters in a square inch, to give its effect. The factor 
(JB2 — jH^)of the formula expressing in it the square of Hues of 
force to the square inch has to be divided by (6.45)^ to give its 
effect for square inch measurement. The value of traction has 
to be divided by 453.6, the number of grams in a pound, to 

reduce it to pounds. This gives a fraction — — rr^-f = 

(6.45)' X 453-6 

This gives for pounds traction and inch measurements, 

292,572 

Traction = — ^^ — ; X = — ^^ ; — ^ • 

24,655 292,572 72,133,627 



194 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. Calculate the traction per square centimeters in 
grams for a field of 350 lines per square centimeter, with an 
induction of 19,820 lines per square centimeter in the case of an 
electro-magnet in contact with its armature. 

Solution. Substituting the values in formula (2) it becomes 
by introducing these values 

Traction = '9>82o^ - 350' _ 392,832,400- i22,.c;oo 
24,655 24,655 

= 16,086 grams. 

For a permanent magnet H is equal to o, and the formulas 
become 

Traction = -r — dynes. (4) 

Traction = — ; — grams. (q) 

24,655^ ^^^ 

Traction = ■ ; — ■ pounds. (6) 

72,133,627^ 

Determination of Permeability from Traction. We have 
seen that B = i^'H. Substituting this value of B in formula (2) 
gives 

Traction ^ ^^^'-°:-^> ^ "-^V-^) . 

24,655 24,655 

whence 

_ \ A4,655 X traction (in grams) 

Example. An iron bar is placed in a field of 5.85 lines to the 
square centimeter. The area of its end is 5 sq. cm. and the 
traction it exerts in contact with another bar in the same field is 
7,995 grams. Calculate the permeability of the iron. 

Solution. Substituting these values in formula (7) gives 

„ _ i A4,655 X 7.995 , ^ 
^ V 5 X (5.85)^ ^ ' 
whence the value of /z, or the permeability of the iron, is 1,073. 



ELECTRO-MAGNETIC INDUCTION 195 



PROBLEMS. 

A piece of iron is placed in a field of 2.4 lines to the square centimeter 
and there are produced in it 129 unit poles. What is its susceptibility? 

^w^- 53-75- 

If 175 units of induced magnetization are produced in a bar of iron 

by a magnetic field of 2.7 intensity, what is the susceptibility of the iron ? 

Ans. 64.8. 

Express the above with conventional symbols. 

Ans. H = 2.7; I = 175; --• = ^ = 64.8. 
-tl 2,7 

If a sample of iron for a value of intensity of magnetization T of 
1,000 has a susceptibility k of 158, what is the intensity of field H 
required to impart such magnetization ? 

Ans. 6.33 Hues of force to the square centimeter. 

A piece of iron of susceptibility 100 is placed in a field of 11.2 lines 
of force to the square centimeter. Calculate its intensity of magneti- 
zation. Ans. 1,120 lines of force to the square centimeter. 

In iron such as that of the table on page 180, what value of S is 
required for a value of J = 574? ^^^ 574 _ 

164 ^'^' 

If the magnetic intensity of a field is 4.7 lines to the square centi- 
meter, and the intensity of magnetization T due to such value of JET is 
875, what is the value of the magnetic induction B? Ans. 11,000. 

What is the susceptibility in the above case? Ans. 186.2. 

Let S" = 585 and let the intensity of the induced magnetization I 
due to it in a piece of iron be 1,530. Calculate the value of S. 

Ans. 19,811. 

Let S = 23,500 and I = 1,600. Calculate the value of S. 

Ans. 43,606. 

What is the value of the permeability m in the last case ? 

Ans. 1.86. 

If the magnetic induction is 15,700 lines to the square centimeter 
and the intensity of the magnetic field producing it is 22 lines of force 
to the square centimeter, what is the permeability of the iron in which 
it is produced? Ans. 713.6 

Calculate the value of /A for -H" = 583, and jB = 19,810. Ans. 33,9. 



196 ELEMENTARY ELECTRICAL CALCULATIONS 

What is the reluctivity of the iron of the last example ? Ans. 0.0295. 

What is the reluctivity corresponding to a permeability of 714? 

Ans. 0.0014. 

If a sample of iron has a permeability of 1,500, what field will be 

required to excite it to an intensity of magnetic induction of ^ = 15,000 ? 

Ans. 10 lines of force to the square centimeter. 

A bar of iron is 112 cm. long and 75 sq. cm. cross-sectional area; its 
permeability at the given induction is 3.5. What are its permeance 
and reluctance? Ans. Permeance 2.34; reluctance 0.428. 

A bar of iron is 44 inches long and 11 sq. inches cross-sectional area, 
and its permeability at the given inductance is 3.5. Calculate its per- 
meance and reluctance. Ans. Permeance 2.22; reluctance 0.45. 

What is the field in the center of a coil 19 cm. long, of 1,015 turns, 
with a current of 0.09 C.G.S. units? 

Ans. 60.4 lines of force to the square centimeter. 

What would be the effect of a current of 0.9 ampere in the same coil ? 

Ans. As 0.9 ampere = 0.09 C.G.S. unit, the result would be the 
same. 

If the above coil was 2 cm. in internal diameter, what would be the 
total number of lines of force produced? Ans. 189.8 lines of force. 

10 amperes are passed through a coil of 275 turns; a bar of iron 39 cm. 
long and 6 sq. cm. cross-sectional area is within it. Assuming that at 
the given induction m = 40, how many lines of force will pass through 
it? Ans. 21,266 lines of force. 

A core of iron is 12 inches long and 3 sq. inches in section. Taking 
/A as 29 and the ampere turns as 750, what will be the magnetic 
induction jB? Ans. 17,346 lines of force. 

What ampere turns are required to force 25,000 lines of force through 
a bar of iron (taking /* as being 2,500), 50 cm. long and 3.22 sq. cm. 
section? Ans. 123.5. 

What ampere turns are needed to force 129,000 lines of force through 
an iron core (taking m at 30) 100 in. long, 12.9 sq. in. section? 

Ans. 10,443. 

An electro-magnet is excited by a field of 4 lines to the square centi- 
meter, causing 9,000 lines of force to pass through each square centi- 
meter of the area of contact between the magnet poles and armature. 
The area of contact is 5 sq. cm. Calculate the traction. 

Ans. 16,422 grams. 



ELECTRO-MAGNETIC INDUCTION 197 

Let the area of contact be 0.775 sq. inch, the field be 25.80 lines of 
force to the square inch, producing an induction of 58,050 lines of force 
to the square inch. Calculate the traction in pounds. 

Ans. 36.2 pounds. 

A field of 15 lines of force to the square centimeter is produced and 
used to excite traction between a magnet and its armature. The area 
of contact is 0.5 sq. cm. and the traction is 4,841 grams. Calculate the 
permeabiHty of the iron. Ans. 1,030. 

The area of contact between the face of an electro-magnet and its 
armature is i square inch; the field is 2,272 lines to the square inch, 
producing a total induction through the area of contact of 19,820 lines 
to the square inch. Calculate the traction in pounds. 

Ans. 5.346 pounds. 



CHAPTER XV. 
CAPACITY AND INDUCTANCE. 

Capacity. — Measure of Capacity. — Capacity of Parallel Plates. — Equa- 
tions for Capacity Calculations. — Energy of a Charge. — Specific 
Inductive Capacity. — Measure of Specific Inductive Capacity. — 
Relation of Absolute Potential, Capacity and Quantity. — Value of 
Absolute Elective Potential. — Potential Difference and Transfer of 
Quantity. — Heat Analogy of Potential and Capacity. — Inductance. 
— The Henry and Rate of Current Change. — Inductance Formulas. 
— Variation of Rate of Current Change. — Energy of the Electro- 
magnetic Field of Force. — Problems. 

Capacity. If a quantity of electricity be imparted to an insu- 
lated conductor it will change its potential. The capacity of a 
body is equal to the number of units of quantity which must be 
given it to change its potential one unit of potential. The poten- 
tial is usually the potential difference between two conductors; 
theoretically the absolute potential may be considered. 

Example. A condenser receives 1 1 units of electricity, caus- 
ing a potential difference of lo® units. Calculate its capacity. 

Solution. If II units of quantity change its potential lo^ units, 

it would evidently require — ^ = ii X io~^units of quantity 

to change its potential one unit. This is the numerical value 
of its capacity. The change in potential will be that which is 
measured between its two sets of leaves. 

Measure of Capacity. The capacity of a conductor or con- 
denser is equal to the quantity of a charge divided by the potential 
which such charge will impart to it. It is numerically equal to 
the reciprocal of the potential which a unit charge will impart. 

Calling quantity Q and capacity K and expressing potential 
difference by y — F', we have 



K = 



Q. 



V 
198 



CAPACITY AND INDUCTANCE 1 99 

Example. A charge of one unit of quantity is imparted to a 
condenser. The potential difference between its plates is thereby 
raised to 12 units. What is its capacity? 

Solution. It is the reciprocal of the potential, ^^ unit of 
capacity. 

The unit of capacity is the farad; the micro-farad is 
generally used in practical computations. (Page 116.) 

Capacity of Parallel Plates. The capacity of conductors of 
different forms generally has to be calculated by the higher 
mathematics. The capacity of two plates facing each other is 
thus determined if one or two assumptions are made. 

Let the potential difference between the two plates be F— F'. 
Assume that the lines of force all proceed straight across from 
plate to plate and that they are evenly distributed. The force of 
attraction between the two plates is 4 rra-, in which or is the surface 
density, so that o-X surface area is the total charge of the surface. 
The potential difference multiplied by the quantity gives the 
energy of the charge; therefore for unit potential the energy is 
numerically equal to the potential difference. The distance from 
surface to surface being t, the energy of the charge per unit of 
surface is 4710- X t, and for unit potential this is equal to F— V\ 

This gives 4 ;r(r X ^ = F - F' (i) and o- = ^~ ^ (2). 

4 7:t 

Let Q represent the charge on one of the plates whose surface 
area is S. Then the charge will be 5 X o- = X S = Q, 

4 7Zt 

The charge on one plate required to impart a potential differ- 
ence of unity between them is the capacity of the condenser. 

O S 
This charge is given by the expression — — — - = K = 

V — V^ 4 Tit 

The capacity is directly proportional to the area and inversely 
proportional to the distance separating the plates. The nearer 
they are to each other the greater will the capacity be. 



200 ELEMENTARY ELECTRICAL CALCULATIONS 

Example. Calculate the capacity of a circular plate i square 
meter (= 10,000 sq. cm.) in area, tV millimeter distant from 
another plate of identical size and shape. 

Solution. Substituting in the formula gives 

^ = . J vToT ^ 79,577 C.G.S. units = 79,577 X lo"* 

4 TT X O.OI 

farads, or 79.6 microfarads. One square meter is equal to 10,000 
sq. cm. and xV millimeter is equal to o.oi cm. 

Equations for Capacity Calculations. The charge of a 
condenser is equal to the capacity multiplied by the e.m.f. due to 
the charge. This gives the equation 

Charge = capacity X e.m.f. (i) 

and by transposing. 

Capacity = ^ (2) and e.m.f. = ?— > (3) 

^ ^ e.m.f. capacity 

which three equations serve for the solution of the simpler prob- 
lems in capacity. 

Example. A condenser has a capacity of 1.3 microfarads, and 
an e.m.f. of 1,325 volts is applied to its terminals. Calculate the 
charge it will receive. 

Solution. By equation (i) the charge is equal to 0.000,001,3 X 
1*325 = 0.001,723 coulomb. 

Example. A charge of 0.000,93 1 coulomb is given a condenser, 
and an e.m.f. of 625 volts is produced between its terminals. 
Calculate the capacity of the condenser. 

Solution. By equation (2) the capacity is ^ X io~® = 

1.49 X io~® = 1.49 microfarad. 

Example. What e.m.f. will a charge of 0.000,725 coulomb 
produce in a 2 -microfarad condenser ? 

Solution. By equation (3) it will be ^-^ = 362.5 volts. The 



CAPACITY AND INDUCTANCE 201 

decimals are omitted because each is of the same order ; it is the 
division of microcoulombs by microfarads. 

Example. A current of 75 amperes average intensity flows for 
^^(X second into a system of 39 microfarads capacity. Calculate 
the e.m.f. developed. 

Solution. 75 X — = 0-375 coulomb. 0.375 "^ 0.000,039 = 
200 

9,615 volts. 

Energy of a Charge. A charged condenser is a seat of poten- 
tial energy. When discharged it delivers electric quantity at a 
definite e.m.f., which constitutes electric energy. As we have 
seen, the charge of a condenser, which is the quantity of electricity 
it can discharge, is equal to the capacity multiplied by the e.m.f. 
This e.m.f. is the maximum e.m.f. due to the charge, and it is 
the initial e.m.f. of the discharge. The final e.m.f. of the dis- 
charge is o, and as the diminution is uniform the average e.m.f. 
of the discharge is one-half the initial. Calling the initial e.m.f. e, 
the quantity discharged is capacity X e; the average e.m.f. of 
discharge is e/2 ; and the energy of discharge is the product of 
the two, or 

Energy of a charged condenser = J capacity X e^. 

Example. What is the energy in a 2.9-microfarad condenser 
charged to 375 volts? 

Solution. Applying the equation we find J X 2.9 X 10-® X 
375^ = 0.204 joule. 

By the doctrine of the conservation of energy the above equa- 
tion gives the energy required to charge a condenser. 

Specific Inductive Capacity. The capacity of two conduc- 
tors facing each other and forming a condenser depends on their 
area, on the distance separating them, and on the material 
between them. This material must be a dielectric or non- 
conductor of electricity. The capacity of a condenser is pro- 
portional to a constant, which constant expresses a property of 



202 ELEMENTARY ELECTRICAL CALCULATIONS 

the dielectric separating its surfaces. This property is called 
the specific inductive capacity of the dielectric. It is also called 
dielectric power, the dielectric constant, permittivity, and per- 
viability. 

Measure of Specific Inductive Capacity. The specific 
inductive capacity of air is taken as unity. That of any other 
dielectric is a number expressing the ratio of the capacity of an 
identical condenser with the other dielectric to that of one with 
air between its surfaces. 

Example. The specific inductive capacity of sulphur is 3.2. 
If a certain air condenser has a capacity of 5 microfarads, what 
would the capacity of an identical condenser with sulphur 
between its surfaces be? 

Solution. It would be 5 X 3.2 = 16 microfarads. 

Example. Taking the specific inductive capacity of glass as 
7, compare the capacities of sulphur and glass condensers, and 
calculate the capacity of a condenser which with sulphur between 
its plates has a capacity of 4 microfarads, when an equal thickness 
of glass is substituted for the sulphur. 

Solution. The ratio is 3.2 -.7=1 :2.i9 (nearly). For the 
condenser in which glass is substituted for sulphur we have the 
proportion 3.2 : 7 : : 4 : x = 8.75. 

Or the capacity of a condenser which with sulphur was 4 micro- 
farads would be increased to 8.75 microfarads by substituting 
glass for sulphur. 

The utility of tables of dielectric constants is of limited value, 
as the values vary greatly with different observers. 

Relation of Absolute Potential, Capacity, and Quantity. 
The absolute electrical potential at a point or place is a math- 
ematical expression with a numerical value. 

Potential expresses a relation between places or loci of force 
such that energy would be required to transfer a quantity from 



CAPACITY AND INDUCTANCE 203 

one place to the other. A level plane, a table top for instance, 
is a locus of force, gravity attracting all objects on it. All parts 
are at the same potential because equally distant from the earth. 
Hence it takes no energy to move a weight from one part of the 
top to another, except for friction and inertia, because there is 
no change of potential. The floor is at a different potential 
because nearer the earth ; hence to move a weight from one to the 
other involves energy. The criterion of the existence of a poten- 
tial difference is the necessary change of energy relations in 
moving a quantity from one place of potential to another. 

Value of Absolute Electric Potential. Absolute electric 
potential is numerically equal to the energy needed to bring a 
unit quantity of electricity from an infinite distance to the point 
where the potential is. This place might be any insulated con- 
ductor. Absolute electric potentials are calculated by the higher 
mathematics. They have little connection with practical electric 
work. Electric potential is equal to the quotient of quantity 
divided by capacity. 

Calling potential V, the formula is 

charge 



V (absolute) or F - V = 



capacity 



2 T 

Example. The capacity of a circular plate of radius r is 

TT 

It receives a charge of 21 units of electric quantity. What is its 
potential, assuming the disk to be 30 cm. in diameter ? 

Solution. 2 r = diameter = 30, and n = 3.1416. Substi- 
tuting in the expression for capacity these values, the capacity of 

the disk is given as — - — - = 9.55 C.G.S. units. The quantity 

.21 
charged upon it divided by the capacity gives potential = = 

2.19 C.G.S. units of electric potential. 
Potential Difference and Transfer of Quantity. The 

difference of potential between any two conductors, or a differ- 



204 ELEMENTARY ELECTRICAL CALCULATIONS 

ence of potential maintained between any two parts of a con- 
ductor, is an actual or possible cause of transfer of electric 
quantity. The transfer is the electric current, and the combina- 
tion of quantity and potential implies and necessitates the 
expenditure of energy. 

The numerical value of potential in any case is equal to that of 
the energy required to transfer a unit quantity of electricity 
against its action. In the case of absolute potential the quan- 
tity is supposed to be transferred from zero potential up to the 
potential stated, so that at the completion of the transfer the 
quantity will have the stated potential. In the case of potential 
difference the quantity is transferred from one to the other 
potential. 

Example. 500 ergs are expended in transferring 29 units of 
electric quantity. What is the potential difference? 

Solution. 500 -r- 29 = 17.24 units of potential. 

Heat Analogy of Potential and Capacity. If we call tem- 
perature of a body its potential and call its specific heat its 
capacity, what has been said of electricity can be illustrated by 
the laws of heat. Thus the energy required to raise a body from 
a temperature m to a temperature n is equal to its mass or quan- 
tity multiplied by its specific heat or thermal capacity. The 
difference of temperatures is equal to the energy divided by the 
mass, exactly as in the last example. 

Example. 500 ergs are expended in heating 29 grams of 
water. Taking the thermal capacity of water as i, calculate 
the temperature or therhial potential which will be imparted to 
the water. 

Solution. As potential difference is numerically equal to the 
energy required to raise unit quantity to the same potential or 
through the same potential difference, the temperature imparted 
to the water will be the quotient of 500 -7- 29 = 17.24° C. This 
is only given as a sort of analogy. 



CAPACITY AND INDUCTANCE 20$ 

Example. The thermal capacity of aluminum being taken as 
0.215, the product of this by any weight will represent the electric 
capacity of a condenser. Let heat imparted to it be called its 
charge and temperature be called potential. If a charge of 
II heat units be imparted to 12 grams of aluminum, what poten- 
tial will be developed ? 

Solution. The formula is V - V = .^llHSL . s^bsti- 

capacity 

tuting for each quantity its value, taking 0.215 X 12 = 2.58 

for capacity, we have 

V-V'=-^ = 4.26° C. 

2.58 

Inductance. The electro-magnetic field of force is a form of 
potential energy. It is produced by passing a current of elec- 
tricity through a conductor. The quantity of energy varies 
according to the conditions and environment of the circuit. To 
create the field the expenditure of energy is required, while it 
is maintained without the expenditure of energy. Inductance 
expresses the conditions and environment of the circuit. 

Suppose that a unit of potential, as a volt, is applied to a circuit 
possessing inductance. Any independent circuit has inductance ; 
one surrounding an iron core, such as an electro-magnet coil, has 
high inductance. A current would be the result, but instead of 
at once taking the value according to Ohm's law it would grow 
in intensity and would build up a field of force. The ultimate 
value of the field of force would be that which the full current 
calculable by Ohm's law would create and maintain, and as soon 
as such field was produced the current would cease to increase 
but would continue at the value calculable by Ohm's law. 

The energy of the current at the given e.m.f . would be expended 
in creating the field until the field attained the value due to the 
current which would be produced in the circuit according to 
Ohm's law. 



206 ELEMENTARY ELECTRICAL CALCULATIONS 

The Henry and Rate of Current Change. The unit of in- 
ductance is the henry. It is the inductance of a circuit which 
requires the maintenance of one volt e.m.f . to increase the current 
passing through it at the rate of one ampere per second. Its 
symbol is L. 

Example. To increase a current passing through a certain 
circuit from o ampere to 39 amperes in 0.001 second 117 volts 
are required. What is the inductance of the circuit? 

Solution. The rate of change is 39 -^ o.ooi = 39,000 amperes 
per second. If 117 volts are required to maintain this rate of 
change, then to maintain a rate of change of i ampere would 
require 117 -^ 39,000 = 0.003 "^olt. The inductance of the 
circuit is 0.003 henry. 

Example. The inductance of a circuit is 2 henrys. What 
e.m.f. must be employed to increase a current through it from 
I ampere to 24 amperes in \ second ? 

Solution. The rate of change is (24— I ) -=- — = 46 amperes 

2 

per second. Multiplying the rate of change by the inductance of 

the circuit gives the e.m.f. required ; 46 X 2 = 92 volts. 

Inductance Formulas. The formulas involved in these 

calculations may be thus expressed : 

Inductance in henrys = 7—7"^ ' (i) 

rate of change 

e.m.f. = inductance X rate of change, (2) 

and a third one may be given, 

Rate of change = . — -^ — — — • (3) 

inductance 

Example. If 21 volts are expended in increasing the current 
through a circuit of 0.004 henry inductance, what is the rate of 
change of current? 

Solution. By formula (3) it is 21 -i- 0.004 = S^^S^ amperes 
per second. 



I 



CAPACITY AND INDUCTANCE 20/ 

Variation of Rate of Current Change. Inductance at the 
instant an e.m.f. begins to act upon an inductive circuit is the 
only opposing factor, because until a current is flowing resistance 
is without effect. The necessary condition of equiHbrium is 
brought about by the creation of counter e.m.f. equal to the 
impressed, and this counter e.m.f., as we have seen, is equal to 
the product of the inductance by the rate of change of current 
intensity. 

After a current has begun to flow a part of the e.m.f. equal to 
the product of this current by the resistance is expended on 
maintaining the current and this amount is the RI drop ; the rest 
of the e.m.f. produces an increase of current whose rate of 
increase or of change is such as to produce a counter e.m.f. 
exactly equal to the residual e.m.f. Knowing the inductance 
and the residual e.m.f. the rate of change is given by formula (3). 

Example. An inductance of 0.04 henry is connected to a 
lighting circuit of no volts. Calculate the rate of change at the 
instant of connection, then at the instant the current has grown 
to 20 amperes, and when it has grown to 25 amperes. The induc- 
tance is of 4 ohms resistance. 

Solution. Applying formula (3) gives as the initial rate of 

change — —=2,750 amperes per second. When the current 
0.04 

has attained a value of 20 amperes, the e.m.f. expended on the 

inductive coil due to its resistance is its RI drop, or 4 X 20 = 80 

volts. This leaves no — 80 =30 volts free to act to increase 

the current and to be in equlibrium with the counter e.m.f. caused 

by such increase. In other words, the current can only increase 

at a rate which will produce counter e.m.f. equal to the e.m.f. 

producing the increase. Applying formula (3) as before, the 



-JO 

rate of change at the 20-ampere point is -^ — = 750 amperes per 

0.04 

second. At the 25-ampere point the RI drop is 4 X 25 = 100 



208 ELEMENTARY ELECTRICAL CALCULATIONS 

volts; the e.m.f. acting to increase the current is no -- loo = 

lo volts, and the rate of change is = 250 amperes per 

second. 
Energy of the Electro- magnetic Field of Force. The 

electro-magnetic field of force is a seat of energy. We have seen 
that the product of inductance by the rate of current change gives 
the initial counter e.m.f. of the field. The counter e.m.f. dimin- 
ishes until the field is fully formed when it is o. The average 

1 1 

counter e.m.f is therefore - L - - This multiplied by the quan- 

2 

tity, or it J the current multiplied by the time, gives the energy 

in electric or equivalent units. Thus we have 

't 
Initial counter e.m.f. =L-, (i) 

L 

1 1 

Average counter e.m.f. = — L - (2) 

2 t 

average counter e.m.f. X quantity = joules, and multipl)dng 
(2) by it we have 

Energy = -W^. (3) 

2 

Example. What is the kinetic energy in a circuit of 0.000,031 
henry through which a current of 29 amperes is maintained ? 
Solution. Applying formula (3) we have 

Energy = - X 31 X lo"^ X (29)^ = 0.013,036 joule. 
2 



PROBLEMS. 

If a charge of 0.2 coulomb give a potential difference of 250 volts 
between the plates of a condenser, what is its capacity? 

Ans. 800 microfarads. 

Calculate the capacity of a pair of circular plates 39 cm. in diameter 
and 0.1 cm. apart. Ans. 951 C.G.S. units; 0.95 microfarads. 



CAPACITY AND INDUCTANCE 209 

charge of 321 microcoulombs gives a potential difference of 750 
volts. What is the capacity of the condenser receiving it? 

Ans. 0,43 microfarads. 

What e.m.f. will 0.001,750 coulomb produce in a microfarad con- 
denser? Ans. 1,750 volts. 

If a 1.5-microfarad condenser is charged to 321 volts, what will the 
measure of the charge be? Ans. 0.000,481 coulomb. 

1. 1 1 amperes flow for yV second into a microfarad condenser. 
W^hat e.m.f. will be developed? Ans. 14,800 volts. 

What is the energy in the charged condenser of the last problem ? 

Ans. 109.52 joules. 

Into the air spaces of an air condenser of 2 microfarads capacity 
parafhne (specific inductive capacity 1.993,6) is poured. What is the 
capacity of the new condenser thus produced? 

Ans. 3.987,2 microfarads. 

Compare the ratios of two condensers of identical dimensions, one 
with sulphur (specific inductive capacity 3.2) and the other with paraf- 
fine as the dielectric. 

Ans. The sulphur condenser has 1.6 times the capacity of the other 
one. 

If sulphur is substituted for paraffine in a condenser of 3. 11 micro- 
farads capacity, what will the capacity of the new condenser be ? 

Ans. 4.98 microfarads. 

2,942,280 ergs are required to transfer 0.333 C.G.S. unit of electric 
quantity from one place to another. What is the difference of potential 
of the two places? Ans. 8,916,000 C.G.S. units; 0.089 volt. 

A current of 3 amperes is to be produced in a circuit of 10 henrys 
inductance in 3 seconds. What e.m.f. will be needed ? Ans. 10 volt. 

An e.m.f. of 200 volts produces a current of 39 amperes by acting on 
a circuit for xV second. What is the inductance of the circuit? 

Ans. 0.466 henry. 

If an e.m.f. of 114 volts acts upon a circuit having an inductance of 
0.57 henry, what will the rate of change be? 

Ans. 200 amperes per second. 



2IO ELEMENTARY ELECTRICAL CALCULATIONS 

An inductance of 0.057 henry is present in a circuit of 3 ohms resist- 
ance; an e.m.f. of 112 volts is maintained on the circuit. Calculate the 
rate of change (a) when the e.m.f. begins to act upon the circuit; (b) 
when the current has attained a strength of 29 amperes; (c) when the 
current has attained a strength of 37 amperes. 

Ans. (a) 1,965 amperes per second. 

(b) 439 amperes per second. 

(c) 17.5 amperes per second. 

An inductance of 0.17 henrys is present in a circuit of 2.9 ohms 
resistance; 1,110 volts e.m.f. is maintained on the circuit. Calculate 
the rate of change (a) at the first application of the e.m.f.; (b) at the 
expiration of the time required for the current to grow to 37 amperes; 
(c) the same for 350 amperes. 

Ans. (a) 6,529 amperes per second. 

(b) 5,898 amperes per second. 

(c) 559 amperes per second. 

What is the energy in the field produced by a current of 3.75 amperes 
passing through a circuit of 0.000,7 henrys? Ans, 0.004,921,875 joule. 



CHAPTER XVI. 
HYSTERESIS AND FOUCAULT CURRENTS. 

Hysteresis Loss. — Steinmetz's Hysteresis Formula and Table. — Stein- 
metz's Formula Based on Weight of Iron. — Foucault or Eddy Cur- 
rents. — Formulas for Laminated Cores. — Formulas for Wire Cores. 

— Copper Loss in Transformers. ■ — Efficiency of Transformers. 

— Ratio of Transformation in Transformers. — Problems. 

Hysteresis Loss. Energy is required to create a field of 
force, but no energy is required to maintain it, and if it is caused 
or allowed to disappear it gives off energy equal to that expended 
on its formation. If the field of force includes a mass of iron 
in its volume or space occupied by it, then there will be a loss 
of useful energy, due to the fact that the iron tends to retain 
some magnetism, and that a small amount of energy is expended 
in its demagnetization which is converted into heat energy 
and operates to increase the temperature of the iron. In many 
cases a transformer would work at an efficiency of nearly loo 
per cent if there were no hysteretic loss. 

The loss due to hysteresis is stated in energy units per cycle 
of magnetization and demagnetization per unit volume of the 
iron per value of magnetic induction (-B), or else in watts per 
unit of weight of iron per value of magnetic induction at a 
stated frequency (cycles per second). The loss may be deter- 
mined experimentally for any given sample of iron, or may be 
calculated by a formula of the empirical class due to Steinmetz. 

Steinmetz's Hysteresis Formula and Table. Let h 
designate the hysteresis loss in ergs per cubic centimeter of the 
iron per cycle, and let V (Greek letter eta) designate a constant 
depending on the quality of the iron. B denotes the magnetic 
induction. Then h = rjB^'^j 

in which h is ergs of hysteresis loss. 



212 ELEMENTARY ELECTRICAL CALCULATIONS 

The following table from Steinmetz gives values of t) for 
different irons. 



Very soft iron wire, 


0.002 


Soft annealed cast steel, 


0.008 


Very thin soft sheet iron, 


0.0024 


Soft machine steel. 


0.0094 


Thin good sheet iron, 


0.003 


Cast steel, 


0,012 


Thick sheet iron, 


0.0033 


Cast iron. 


0.0162 


Most ordinary sheet iron) 
for transformer cores, ) 


0.004 
to 0.0045 


Hardened cast steel, 


0.025 



Example. The iron in a transformer core is subjected to 
8,000 lines of force per square centimeter at the extremes of 
the cycle. Calculate the loss by hysteresis per cycle. 

Solution. Substituting in the formula we have 
h = 0.0045 X (8,ooo)^"^ ergs. 
The logarithm of 8,000 is 3.903,090. Multiplying it by 1.6 
gives the logarithm of (8,ooo)^'®, which is 6.244,944, the number 
corresponding to which is 17,577 X lo^ This gives as the 
value of h 

0.0045 X 17,577 X io2 = 7,910 ergs, 
the hysteretic loss in iron of that quality (jj = 0.0045) P^r cycle 
per cubic centimeter with a maximum induction of 8,000. 

Steinmetz's Formula based on Weight of Iron. Iron 
weighs about 7.7 grams per cubic centimeter and a pound is 
equal to 453.6 grams. There are therefore 453.6 -v- 7.7 = 58.9 
cubic cm. in a pound of iron. Call the number of cycles per 
second /", and the formula becomes for one pound and for 
/ cycles per second, in ergs per second, 

and as 10^ ergs per second are a watt, the formula becomes for 
watts of hysteresis loss and for n pounds 

W = 58.9 X 10-^ fvB^'^ X n. 
Example. A two-pole dynamo armature makes 15 revolu- 
tions per second. It weighs 27 pounds. The average number 



HYSTERESIS AND FOUCAULT CURRENTS 21 3 

of lines of force per square centimeter of cross-sectional area of 
field is 12,000. Let the iron have a hysteresis factor of 0.003. 
Calculate the rate of loss. 
Solution. Substituting in the formula we have 

W = 58.9 X 10-7 X 15 X 0.003 X i2,oooi-« X 27, 

which gives as solution W = 24 watts. This is the rate at which 
energy will be absorbed, expending itself in heating the arma- 
ture core. The answer of the formula can be equally well 
read in joules per second. 

Within reasonably high values of B this formula gives results 
not over 3 per cent different from those obtained by experiment. 

Foucault or Eddy Currents. The name Foucault currents 
or eddy currents is used to designate currents produced in the 
cores or other masses of metal of electrical machinery by the 
variations in the magnetic induction to which they are sub- 
jected in the operation of the machines. These currents do no 
direct injury unless they become so intense as to overheat the 
metal so as to affect the insulation, but are indirectly harm- 
ful as absorbing energy and thereby making the operation of 
the machinery less efficient, so that they act with hysteresis to 
produce a waste of energy. 

Formulas for Laminated Cores. When a core is built up of 
sheets of iron insulated from one another the following formula 
expresses the loss in watts or in joules per second. In it 
/ = frequency. W= watts absorbed per cubic cm. 

c = specific conductivity. JB = magnetic induction. 
a = thickness of plates. 
We then have 

4 X 10^^ 
There is another factor in the formula which is omitted because 
it is practically equal to unity. The value of c will vary widely 



214 ELEMENTARY ELECTRICAL CALCULATIONS 

for different irons. For such as are used in converters and in 
armatures a high value is 1.02 X 10^; this is at the temperature 
of zero; for each degree increase of temperature a decrease of 
conductivity of 0.365 per cent may be allowed for. This is 
applied to the value of c in the formula before substituting. 

The value of c in the formula is in the reciprocal of ohms per 
cubic centimeter ; the value of a is in centimeters ; the value of 
B is in lines of force per square centimeter. 

Example. A core is built up of plates of iron 0.2 cm. thick 
and of the conductivity 1.02 X 10^ at zero C. In operation it 
becomes heated to the temperature of 200 degrees C. The 
frequency of alternations is 12 per second. The induction is 
10,000 = 10^ lines of force to the square centimeter. Calculate 
the loss due to eddy currents. 

Solution. The conductivity at 20 degrees C. is [1.02 — (1.02 
X 0.00365 X 20)] X lo^ = (1.02 - 0.07) X lo^ = 0.95 X lo^ 
Substituting this and the other values in the formula we obtain 

4 X lO^^ 

_ 9.87 X 144 X 10^ X o 04 X 0.95 X 10^ 
"" 4 X lO^® 

= 9.87 X 1.44 X 0.95 X io~^ = 0.0135 watt. 

This is the rate of loss per cubic centimeter of the core. The 
insulation is not to be included in the volume. 
Separating the formula thus : 

W = X pB'^a'^j and assigning an average value to c, we 

may express the fraction, ' as a constant. Calling this 

constant J, the formula becomes 

W^ hfBH\ 



HYSTERESIS AND FOUCAULT CURRENTS 215 

To show how h is deduced assume the conditions of conductiv- 
ity of the last problem. We then have as the value of 5, 

, 9.87X0.9^X10^ ^/ _ii 

^- 4X10-"— = '-^44X10" 

and the formula becomes 

W= 2.344 X io~" X 144 X 10^ X 0.04 = 0.0135 as before. 

This is too high a value of h to represent ordinary practice. 
An accepted value is 1.6 X lo"^^ As the value of h varies with 
every sample of iron and with every change of temperature, it is 
evident that the formula based on a constant value of h must be 
an approximate one only. Unless the temperature and con- 
ductivity of the iron are accurately known, the original formula 
will also be an approximate one. 

Formulas for Wire Cores. If a core is made of wire, the 
formula must include the diameter of the wire in place of 
the thickness of the plates. For wire cores the formula is 
the following: 

16 X ioi« 

7^ y. c 

If the constant is to be used, it is equal to ' which 

^ 16 X 10^^ 

is the same expression as already used for the value of &, except 
that 16 in the divisor replaces 4; therefore any value of h that is 
applicable to the formula for laminated cores becomes applicable 
to wire cores by dividing it by 4. Thus for a wire core, instead 
of 1.6 X io~^^ we would have to use 4 X 10-^^, which is one- 
fourth of the value of h for plate cores. Calling this constant 
for wire cores V we have, as the formula embodying the con- 
stant h' for wire cores. 

Example. A core is made of iron wire insulated as usual. 
The frequency is 24, the induction is 12,000 lines of force 
to the square centimeter, the conductivity of the iron at zero 



2l6 ELEMENTARY ELECTRICAL CALCULATIONS 

is 0.78 X 10^, the temperature of operation is 36 degrees C, 
and the wire is 0.3 cm. diameter. Calculate the eddy currents 
loss per cubic centimeter. 

Solution. This statement calls for the application of the orig- 
inal formula for wire cores. The value of c is deduced as 
before; it is [0.78 -(0.78 X 0.00365 X 36)] X 10^ = o.68Xio^ 
Substituting this and the other values of the problem in the 
formula we have 

.„_ 9.87 X (24)^X0.09 X (12 X ipy X 0.68 X 10^ 
16 X ioi« 
_ 9.87 X 576 X 0.09 X 144 X 0.68 X 10^^ _ 
16 X 10 



16 - 0.0313. 



This is the value of the loss due to eddy currents per cubic 
centimeter of iron, in rate units (watts) or in joules per second. 

Example. Make the same calculation but using the con- 
stant of the problem before modified so as to apply to wire. 

Solution. The constant is 1.6 X lo-^^ For it to apply to 
wire it must be divided by 4, giving 4 X io~^^. Substituting in 
the formula W = VpBH\ 

W= 4 X 10-12 X (24)2 X (12 X 10^)2 X 0.09 = 0.0299, 

which differs from the other solution on account of the constant c 
having a value sUghtly different from that assigned it in the first 
treatment of the problem. 

It has been claimed that in all these problems the induction B 
should be raised to the i .6 power instead of to the square. 

Copper Loss in Transformers. The sources of loss in trans- 
formers are hysteresis, eddy currents, and copper loss. The 
former two have already been treated ; the copper loss is simply 
watts expended in the primary and secondary coils, which are 
equal to the product of current by e.m.f. absorbed by each of 
the coils or to the resistance of each of the coils multiplied by 
the square of the current. 



HYSTERESIS AND FOUCAULT CURRENTS 217 

Denoting the primary current and resistance by /j and R^, 
and denoting the same for the secondary by the same letters with 
subscript 2, the copper loss is given by the following formula: 

Per cent of copper loss = : ' . ^^-^^ X 100. 

input in watts 

The multiplication by 100 is necessary to give the per cent 
factor; without such multiplication the result is a decimal frac- 
tion, which can be used equally well. 

Example. A primary circuit delivers 3,200 watts to a trans- 
former. The resistance of its primary is 3 ohms, the resistance 
of its secondary is 300 ohms. What is the per cent of copper 
loss for a primary current of 1.7 amperes and a secondary cur- 
rent of 0.8 ampere. 

Solution. Substituting in the formula gives 

Per cent of copper loss = ^ X Ml + Lsoo X (0.8)^] ^ ^^ 

3,200 

200.67 , „ 

= ^ X 100 = 6A^ per cent. 

3,200 

Efficiency of Transformers. The efficiency of a trans- 
former is equal to the output divided by the input. If to be 
in per cents the quotient is multiplied by 100. Otherwise it 
will be in simple decimal notation. The output is equal to 
the input diminished by the sum of the eddy current loss, the 
hysteresis loss, and the copper loss. The constituents of this 
expression are all in watts. The efficiency is 

Per cent of effi. = fap"t-(ed.cu r .loss+hys.loss+cop.loss) ^ ^^^ 

input 

Example. What is the efiiciency of a transformer in which 
the eddy current loss is 52 watts, the hysteresis loss is 47 watts, 
and the copper loss is 62 watts, at an input of 11 horse -power ? 

Solution. 1 1 horse-power = 746 X 11 = 8,206 watts. Apply- 
ing the formula we have 



2l8 ELEMENTARY ELECTRICAL CALCULATIONS 

Per cent of efficiency = 8.^°6 -(52 + 47 +62) ^ ^^ 

8,206 
_ 8,045 
~ 8,206 ^ ^°° ^ ^^-^"^ P^^ ^^^^• 

This is about the maximum efficiency attained in practice. 

Ratio of Transformation in Transformers. The ratio of 

transformation in a transformer is the ratio of the impressed 

e.m.f. to the e.m.f. delivered to the secondary. The ratio is 

equal to the turns in the secondary coil divided by the turns in 

the primary coil. It can be expressed thus : 

r ^ J. r r ' v, tums iu sccondary 

e.m.f. output = e.m.f. of primary X : : ^ • 

turns m primary 

Example. A transformer is actuated by a voltage of 1,500 

volts. The primary coil has 1,000 turns, the secondary has ^^ 

turns. Calculate the e.m.f. delivered. 

Solution. By the formula it is equal to 1,500 X — ^ = 49.5 

1,000 

volts. 

Example. What must the ratio of primary to secondary be to 
transform 2,000 volts to 60 ? 

Solution. It must be 2,000 : 60 = 33 J : i. 

PROBLEMS. 

In a field of 9,500 lines of force to the square centimeter at the maxi- 
mum, with a hysteretic constant of 0.0045, calculate the loss of energy 
per cycle. Ans. 10,413 ergs. 

With a maximum of 65,000 lines of force to the square inch, and with 
a core of 0.0043 hysteretic constant, what is the loss per cubic centimeter 
per cycle? ^ Ans. 10,935 ergs. 

A core weighs 357pounds; the polarity changes 2,500 times a minute; 
the constant of, the core is 0.0024, with a maximum field of 114,000 
lines to the square inch. Calculate the rate of loss. 

Ans. 1,313.8 watts. 

A core weighs 250 pounds; there are 16 cycles per second; the maxi- 
mum field is 21,000 lines to the square centimeter; the hysteretic con- 
stant is 0.003. Calculate the rate of loss. Ans. 581.9 watts. 



HYSTERESIS AND FOUCAULT CURRENTS 219 

The plates of a laminated core are 0.3 cm. thick; the iron of the core 
has a specific resistance of 1.04 X 10* at 0° C, with a coefl&cient of 
0.00365 increase in resistance per degree C; in operation it is heated 
to 25° C; the frequency is 17 per second; the maximum induction is 
15,000 to the square centimeter. What is the loss per cubic centimeter 
due to eddy currents? Ans. 0.1359 watt. 

With a value for b in the short formula of 1.6 X lo"", a maximum 
induction of 14,780 lines of force to the square centimeter, frequency 
16 per second, and a thickness of laminations of 0.27 cm., calculate the 
loss per cubic centimeter. Ans. 0.06523 watt. 

A core is made of iron wire of o.i cm. thickness and carries a maxi- 
mum induction of 23,500 lines of force to the square centimeter; the 
frequency is 25 per second; the conductivity of the wire is 0.78 X 10^ at 
0° C. ; the temperature of operation is 39° C. Calculate the loss per 
cubic centimeter. Ans. 0.014 14 watt. 

Apply the short formula, taking the thickness of the wire at 0.13 cm., 
the frequency at 25 per second, the maximum induction at 23,400 lines 
per square centimeter, and a value of 4 X io~^^ for 6'. Calculate the 
loss per cubic centimeter. Ans. 0.2313 watt. 

The primary coil of a transformer has 900 turns and receives a volt- 
age of i,2Qo. How many turns must there be in a secondary coil to give 
16 volts? Ans. 12 turns. 

A primary circuit delivers 3,100 watts to a transformer the resistance 
of whose primary and secondary coils are 60 ohms and 5 ohms 
respectively. If there is a current of 0.4 ampere in the primary coil 
and 3 amperes in the secondary, calculate the per cent of copper loss. 

Ans. iHI per cent. 

The same transformer has a hysteresis loss of 32.4 watts and an 
eddy current loss of 38.2 watts. Calculate the per cent of efi&ciency. 
Ans. 95Hf per cent, or 96 per cent nearly. 



CHAPTER XVII. 
ALTERNATING CURRENT. 

Induction of Alternating E.M.F. — Alternating Current. — The Sine 
Curve. — Sine Functions. — Cycle. Frequency. — Value of Instan- 
taneous E.M.F. and Current. — Average Value of Sine Functions. — 
Eflfective Values. — Form Factor. — Reactance of Inductance. — Rate 
of Change . — Deduction of Ohmic Value of Inductance Reactance. — 
Impedance of Inductance and Resistance. — Lag and Lead. — Lag of 
Current. — Deduction of Ohmic Value of Capacity Reactance. — 
Combined Impedance of Inductance and Capacity. — Lead of Cur- 
rent. — Lag or Lead due to both Reactances. — Impedance of Induc- 
tance, Capacity, and Resistance Combined. — Angle of Lead or Lag. 
— Power and Power Factor. — Power Factor for both Reactances 
Combined. — Angle of Lag and Rate of Change. 

Induction of Alternating E.M.F. If a coil of wire 
rotates in a uniform field, the coil representing practically the 
circumference of a disk, electro-motive force will be generated in 
it. This e.m.f. will vary in amount and in polarity. Assume 
the coil at right angles to the field and turning. At that point 
no e.m.f. will be impressed upon it, because it cuts no lines of 
force. As it rotates it begins to cut lines at a rate increasing 
until a maximum rate is reached at the position when its plane 
is parallel with the lines of the field. Here the e.m.f. is at its 
maximum, and as it passes this position the e.m.f. begins to fall 
off, because the coil cuts the lines at a lower rate, until as it 
reaches the position at right angles to the lines of the field the 
e.m.f. falls again to zero value. So far the e.m.f. has risen and 
fallen, producing a field of one polarity, but as the coil in its rota- 
tion begins to move through the other half of its course the 
e.m.f. goes through exactly the same set of changes as before, 
except that the polarity of the field produced is the reverse of 
what it was before. 

Alternating Current. If the ends of the coil are connected 
by a conductor a current will pass through it which will vary 



1 



ALTERNATING CURRENT 



221 



in intensity by the same law regulating the e.m.f., and which 
will therefore vary from a maximum to zero, and which will go 
alternately in opposite directions as the polarity of the e.m.f. 
changes. 

Such a current is called an alternating current, and the division 
of the science relating thereto is called alternating current elec- 
tricity. 

The Sine Curve. A current produced by the arrangement 
described is what is known as a sine current. If a horizontal line 
is drawn and is taken to represent the development of the 
circumference of a circle, called the generating circle, any given 




length laid out or measured off on the line may be taken as 
representing an angular quantity, measured in radians or 
degrees. The entire line will represent 2 re radians, or 360 
degrees. Erect on the first half of the line a number of ver- 
ticals each of which will be at a point referable to angular 
measurement, as at the lo-degree mark, 20-degree mark, and 
so on. For the second half of the line, as the polarity of the 
e.m.f. is supposed to reverse itself at the center of the line, 
other lines are erected, but are inverted in position, or directed 
downwards. Assume the radius of the generating circle to be i . 
Each line is made of length equal to the sine of the angle repre- 
sented by its position. The line on the lo-degree mark, for 
instance, would be 0.17365 high, and the lines on the 170- 
degree, 190-degree, and 350-degree marks would be of the 



222 ELEMENTARY ELECTRICAL CALCULATIONS 

same lengths, because these are the sines of the angles named. 
The last two lines would be directed downwards. In this way 
any number of lines could be laid out and the relative lengths 
of the lines made to vary as the lengths of the sines of the angles 
to which the respective Hnes belong. The values of the sines 
can be taken from a table of natural sines. A line drawn 
through the ends of the perpendiculars takes the form of a 
wave; it represents a cycle or wave of current or of e.m.f. and 
is a sine curve. 

With the exception of the values for the 90-degree and 270- 
degree points which are equal to the radius, because the sine of 
either of these angles is i, the values of the lines, which are 
called ordinates of the curve, will be expressed as decimals of 
the radius, if the radius is taken as i, for the basis of the 
decimals is the radius of the generating circle. This radius 
is equal to the quotient of the length of the horizontal line, 
which is called the abscissa of the curve, divided by 2 tt, which 
is 6.2832. 

Example. A sine curve is to be laid out on a line 4.25 in. long. 
Where should the lo-degree ordinate be erected and what should 
its length be? 

Solution. 10° is — - of the line. 4.25 X — — = o.iiSin., 

360 360 

which is the distance from the end at which the ordinate should be 

drawn. The radius of the generating circle is ' = 0.677. 

6.2832 

From a table of natural sines we find that the sine of 10 degrees is 

0.17365. The length of tlie lo-degree ordinate is the product 

of the radius of the generating circle by this decimal, or 0.677 X 

0.17365 = 0.1175^ inch. As the sines of 10 degrees, 170 degrees, 

190 degrees, and 350 degrees are equal, this is the length of the 

four ordinates at the four degree points stated. 

Sine Functions. The vertical Unes or ordinates erected on 

the base line as described, and whose lengths are proportional to 



ALTERNATING CURRENT 223 

the values of current and e.m.f. at the angular distances indi- 
cated, are sines of the angles, consequently the currents and 
e.m.f. 's are called sine functions. 

Cycle. Frequency. A complete cycle starting from zero 
increases to a maximum in one direction, returns to zero, 
increases to a maximum in the other direction, and returns 
to zero. If drawn as a sine curve it starts from the base line, 
rises to a crest and returns to the base line, and then accom- 
plishes the equivalent below the base line. The number of 
times this action takes place in a second, or the number of 
cycles produced in a second, is the frequency of the circuit, 
of the current, or of the e.m.f. 

Value of Instantaneous e.m.f. and Current. In the ordi- 
nary operation of alternating current systems the polarity of the 
e.m.f. reverses many times in a second. The e.m.f. and the 
resulting current, which reverses in direction exactly as the e.m.f. 
reverses in polarity, are indicated by two sine curves, often of 
different altitudes but necessarily of the same angular length. 
Under some conditions the two curves may coincide so as to be 
one. Knowing the value of the maximum e.m.f. of current, the 
value of the e.m.f. or current at any part of the cycle defined by 
its angular position can be stated in terms of Ejj^g^^(E maximum), 
the maximum e.m.f. or I^^^^the maximum current. CalHng the 
e.m.f. at any part of the cycle e, its value is given by the equation 
e = ^max sin d, and the equivalent process gives the current 
as i = Ij^^ sin d, indicating the angular position of the sine 
function. Such are termed instantaneous values. 

Example. An alternating current generator produces a maxi- 
mum e.m.f. of 2,000 volts. What is the e.m.f. of the cycle when 
f completed? 

Solution. 360° X - = 154° 17'. This is the angle 6) its 
7 
sine is 0.43392, and substituting in the formula, 

g =2,000X0.43392 =867.84 volts. 



224 ELEMENTARY ELECTRICAL CALCULATIONS 

Average Value of Sine Functions. The average value 
of the e.m.f, or of the current of an alternating current system 
may be calculated approximately by adding together a number 
of the values of evenly distributed sines proportional to the 
e.m.f. 's or currents for every five or ten degrees of the wave 
length, dividing by the number taken, and multiplying the result 
by the maximum e.m.f. or current. It can be done with one- 
quarter of a wave, which is 90 degrees. 

Example. Calculate the average value of the e.m.f. of a sine 
wave whose maximum e.m.f. is 2,250. 

Solution. From a table of natural sines we obtain the following 
values of the sines of o degrees, 10 degrees, 20 degrees, and so on 
up to 90 degrees: 0., 0.17365, 0.34202, 0.50000, 0.64279, 
0.76604, 0.86603, 0.93969, 0.98481, 1. 0000. Adding these 
together and dividing by 10, the number of values, we have 
6.21503 4- 10 =0.6215. 

Approximate average value of the sines =0.6215 and multiply- 
ing by the maximum e.m.f. gives 2,250 X 0.6215 = 1,398.4 volts. 

The result is only approximate, the true average being 1,432 
volts. The result could have been made more accurate by taking 
more sine values, as one for every 5 degrees or for every 2 degrees. 

The length of one-half of the wave length is n Emax in 
which £max is the length of the radius of the generating circle. 
If we know the area of the space included between the horizontal 
base Une and the half of the sine curve lying above or below it, 
it is evident that the quotient of the area divided by the length 
of the half of the base line in question will be the average length 
of the vertical Hues representing the sine functions. By cal- 
culus the area of the half of the sine curve is determined; it is 
2 £^max. For the value of the average e.m.f. this is to be divided 
by the length of the base line of the half of the curve, the value 

of which length is given above. This gives 
2 £2 2 

Average e.m.f. = °°^^ = - E^^^= 0.6366 Ej^^x- 



ALTERNATING CURRENT 225 

Example. What is the average value of the e.m.f. of an 
alternating current system whose maximum e.m.f. is 1,200 
volts? 

Solution. It is 1,200 X 0.6366 = 764 volts. 

If e = — .£max, as above, then £max = — e = 1.57 e. By this 

TZ 2 

formula if e is given £max can be calculated. 

Example. The average value of the e.m.f. of an alternating 
circuit being 875, what is the maximum e.m.f.? 

Solution. It is 875 X 1.57 = 1,373-75 volts. 

Effective Values. With a constant resistance the energy 
of an active circuit varies with the square of the electro-motive 
force, the expression for energy being £^/i?, and also with the 
square of the current, the expression for energy being RP. 
If one of these is true the other must also be true, because by 
Ohm's law the current varies with the e.m.f. Either expres- 
sion reduces to E/, or, in practical units, to watts, the unit of 
energy rate. It follows that an alternating current or an alter- 
nating e.m.f. will represent at any instant a heating or energy 
effect proportional to the square of its value at that instant. 
The average effect will vary with the average of these 
squares, and this average will be the value of a direct current 
or direct e.m.f. of the same energy effect. This energy relation 
of alternating currents and e.m.f.'s is their virtual or effective 
relation, and for an alternating circuit to have the same energy 
as a direct circuit of the same elements the square of its effect- 
ive current or e.m.f. must equal that of the same element of 
the direct circuit. The relation of the currents or e.m.f.'s 
unsquared, then, is that of the square root of these squares. For 
the direct circuit this gives the original current and e.m.f.; for 
the alternating circuit it gives the square root of the average 
square. If the square root of the average square of an alter- 
nating current is equal to a direct current, the energy developed 



226 ELEMENTARY ELECTRICAL CALCULATIONS 

in the passage of either one through the same resistance will be 
the same. The same applies to the expenditure of e.m.f. 

Effective values of current or e.m.f. are indicated thus: /^^ 

or Eef. 

The square root of the average square constitutes the effect- 
ive or virtual current or e.m.f., I^^ or E^j. Its value is thus 
determined : A curve is laid off similar to a sine curve and on 
the same base line, but the verticals or ordinates, instead of 
varying with the sines, are laid off equal to the squares of the 
sines. The area of the space thus inclosed is determined by 

calculus ; it is equal to ^^^- If this area is divided as before 

2 

by the length of the base line, n Emax, it will obviously give a 
quotient which is the average of the squares of the sines; this 

, £ max , , . , . , 

quotient is j and its square root is the square root of the 

Tj' 

average squares and is equal to -^^ = £ef» whence E^^^ = 
\/~2 Eqi. The square root of 2 is 1.414; — — is 0.707 ; so E^f = 

\/2 

0.707 £max, and £max = 1.414-Eef- The same formulas apply 
to the effective current values. 

Example. What is the effective value of an alternating 
current whose greatest value is 31 amperes ? 

Solution. It is 0.707 X 31 =21.9 amperes. 

The relation of the maximum current or e.m.f. to the effective 
may be thus proved by trigonometry. 

The law of the sine curve applied to current strength gives the 
equation, letting / stand for any instantaneous value of current, 
in this case at the point d, 

/=/^,,sin^, (I) 

and squaring, 

P = P^sm'd. (2) 



ALTERNATING CURRENT 22/ 

The sum of the squares of the sine and cosine of any angle is 
equal to i. The sines and cosines vary in value by exactly the 
same law, so for the average functions we have 

Average sin^^ + average cos^^ = i, (3) 

and as the average sine is equal to the average cosine, we can 
substitute one for the other, and doing this we have 

2 average sin^^ = i (4) 



and 



and 



Average sin^^ = - (5) 



Vaveragesin^ =\/- =0.707. (6) 

If in equation (i) we substitute the square root of average 
sin^^ for sin 6, the second member of the equation will be the 
average value of the effective current, because the result will 
be the square root of the average values of the squares of the 
currents, thus: t _ ^ ^^« r /-\ 

^ef = 0-707 ^max- (7) 

Form Factor. The quotient of the effective value of an 
alternating current or e.m.f. divided by the average value is the 

form factor. For the sine curve current this is equal to -^~^ = 

0.636 

I. II. 

As alternating systems are generally operated on sine curves 
the form factor in practice is taken as i.ii. 

Reactance of Inductance. When a current varies in strength 
it causes a change in the electro-magnetic field of force which 
always surrounds a current. This change is one of strength of 
field. As the current increases it builds up the field, producing 
more lines of force ; as the current decreases the field diminishes 
in strength also, the lines disappearing. Thus if a current is 
normally an increasing one its increase will be opposed by the 



228 ELEMENTARY ELECTRICAL CALCULATIONS 

above action, because energy has to be expended to build up a 
field. If it tends to decrease, the tendency to decrease will be 
opposed also, because energy is given off in the reduction of a 
field, and this action tends to increase a current in the original 
direction or to diminish the rate of decrease of a decreasing cur- 
rent which produced or maintained the field. 

The effect of inductance as outlined above is then to oppose 
the normal changes or normal action of an alternating current. 
This effect is called reactance. Its value depends on the rate of 
change of the current and on the inductance of the circuit. 
To put the effect of inductance into a form available for calcu- 
lation, it can be expressed as equivalent to a definite number of 
ohms. 

Rate of Change. The rate of change of an alternating current 
at any point expressed in degrees is equal to the product of the 
maximum current by the frequency by the cosine of the angle of 
position ^ by 2 TT. The equation is 

Rate of change = 2 Ttfl^^^ cos d. 

The numerical value of the rate of change is independent of its 
positive or negative sign, so that the sign of cos 6 is disregarded. 

Example. What will be the rate of change in an alternating 
current of 133 frequency, effective value 65 amperes, at 180 
degrees? 

Solution. The maximum current is equal to 65 X J. 414 = 
92 amperes. The cos 180 degrees is — i. Substituting in the 
formula and disregarding the sign we have 

Rate of change = 92 X 2 ;r X 133 = 76,881 amperes. 

Example. In a sine current where is the maximum rate of 
change? 

Solution. The variable in the equation is cos 6. The points 
where this has its highest values are the points of maximum rate 
of change. From trigonometry we know that the cosines of 



ALTERNATING CURRENT 229 

o degrees, 180 degrees, and 360 degrees are the highest in value; 
therefore the rate of change is highest at these points. The 
minimum rate is at the 90-degree and 270-degree points, where 
the cosines are of zero value, and consequently the rate of change 
at these points is o. 

Deduction of Ohmic Value of Inductance Reactance. 
The rate of change of the current at any point d is given by the 
expression 2 nflj^^^cos 6. The period of greatest rate of change 
is that at which cos d has the greatest value, and the maximum 
value of a cosine is when the arc has a value of o degrees or of 
180 degrees; its value is then i. The e.m.f. due to inductance 
is equal to the product of the rate of change by the inductance. 
Calling the inductance Z,, the e.m.f. due to it is equal to 
2 tt/L/^j^^ at the point of maximum value. By Ohm's law 
the e.m.f. is equal to RIj^^^ for a current /^^xj ^^^ ^^'^ hsive 
2 TifLI^^^^ = -R/jjiax' whence R = 2 nfL. Therefore the ohmic 
equivalent of the inductance of an alternating circuit is equal 
to 2 nfL. 

This gives the formula 

Reactance = 2 tt/L, (i) 

in which L is the inductance of the circuit in henrys, and / is 
the frequency of the current. The value is given in equivalent 
ohms. 

Example. A coil of wire is of such inductance that a current 
changing at the rate of one ampere a second induces a counter 
e.m.f. of 0.025 volt. An alternating current changing 100 times 
a second passes through it. Omitting any consideration of the 
true ohmic resistance, or assuming that it is so small as to be 
negligible, what is the ohmic equivalent of the reactance? 

Solution. From the data of the problem, L = 0.25, / = 100, 
and we have 

Reactance = 2 tt X 100 X 0.025 = 15.7 ohms (equivalent). 



230 ELEMENTARY ELECTRICAL CALCULATIONS 

The frequency of a current is the number of periods or waves 
per second in its operation. If T is the time of a period, then 
the frequency of the current is obtained by dividing i second 

by the time of a period ; or F = / = — > which expression intro- 
duced into the reactance formula, as given above, makes it read 
Reactance = — — - • (2) 

The formulas are exactly equivalent, and some authors use 
one and some the other. 

Example. In a circuit of 0.051 henry an alternating current 
is produced with a period value of yys second. Calculate 
the reactance. 

Solution. Substituting in the last formula, 

2 ttL 

Reactance = — ;=- = 56 ohms (equivalent). 

Angular velocity is equal to angle traversed in a second. If 
expressed in radians, and if T is the time required to traverse 
an angle of the value 2 tt, then such angular velocity has the 

value of — • Angular velocity being denoted by o) (Greek 

letter omega) we have 

^ = ^' (3) 

2 TtL 

and substituting this in the equation. Reactance = —;=;- gives a 
third form of the reactance equation which is often used, 

Reactance = Luj. (3) 

Example. A current has a frequency of 133 and is passing 
through a circuit of 0.090 henry. Calculate the reactance by 
the angular velocity formula. 

Solution. The time required for the completion of a period 
or wave is ri^ second, and the angular length of a period is 



ALTERNATING CURRENT 23 1 

2 n, or 6.2832 radians. The angular velocity of the current is 
therefore 6.2832 -v- ' — = 835.66 radians. Substituting this 
value in the reactance formula we have 

Reactance = 0.090 X 835.66 = 75.21 ohms (equivalent). 

For general purposes the formula (i), using frequency 
directly, is the most convenient. * Equivalent ' is often 
omitted, but is then to be understood. 

Impedance of Inductance and Resistance. Reactance 
and resistance act together in an alternating current circuit to 
reduce a current due to a given e.m.f. Their combined action 
is equal to the square root of the sum of their squares, and 
their combined action is called impedance. Using the value of 
reactance of equation (i) we have 



Impedance = Vr^ + (2 7:fL)\ 
and equations (2) and (3) would give for the same the values 



\M¥J 



and Vr^ + (Lcoy. 



Example. A coil of wire has a resistance of 23 ohms and an 
inductance of 0.02 1 henry. What is its impedance for a current 
of frequency no? 

Solution. 2 tt/L = 6.2832 X no X 0.021 = 14.5. Then 

V(i4.5)^ + (23)^= V739-25 = 27.2 ohms, 
which is the impedance of the circuit for a current of the given 
frequency. 

Lag and Lead. If the ends of the coil we have described 
are connected, the current due to the e.m.f. impressed on it will 
take the form of waves of exactly the angular length of the e.m.f. 
waves and of the same form. The current waves may corre- 
spond in position with the e.m.f. waves — crest lying over crest 
and both sets of waves crossing the base line at the same point — 



232 ELEMENTARY ELECTRICAL CALCULATIONS 

or the two sets may vary in position. One set may reach the 
crest before the other does; the one in arrears is then said to 
lag. The amount of its lag is measured on the base line and is 
referred to the angular measurement of this line or, what is the 
same thing, to that of the generating circle. The lag is stated 
in degrees generally, sometimes in radian measurement. The set 
of waves in advance of the other is said to lead, and its position 
is stated, in angular measurement also, as the angle of lead. 

If the height of the waves is used to indicate the measurement 
of current or of e.m.f., the two sets may vary in height. As a 
matter of convenience and to distinguish the two sets from one 
another they are often drawn of different heights. 

Example. A set of current waves cross the base line yV of 
its length behind or later than the e.m.f. waves. What is the 
angle of lag ? 

Solution. The length of the line in angular measurement is 

360 degrees, and 360 X — = 36 degrees, which is the angle of 
10 

lag. To express it in radian measurement multiply the length 

of the base line in radians, 2 tt or 6.2832 radians, by — > giving 

10 

0.62832 radian as the angle of lag. As a radian is equal to 

57.3 degrees, as a test of the correctness of the operations we 

may multiply it by 0.62832 and see if it gives the first result, 

36 degrees. 57.3 X 0.62832 = 36.002 degrees, which is well 

within the Hmits of the decimals used. 

Lag of Current. This is produced by inductance ; the angle 

of lag, indicated by <^ (Greek letter phi), is the angle whose 

tangent is equal to the quotient of the reactance of induction 

divided by the resistance. This gives the equation 

_ ^ . 1 r 1 . / reactance 2 rrfL 

Tangent of angle of lag = tan 0= — — = — ±-' 

resistance i\ 

_ 2 TiL _ Loj 



ALTERNATING CURRENT 233 

Example. A circuit through which a current is passing has a 
resistance of 3 ohms, and with the current in question going 
through it has a reactance of 5 ohms. What is the lag of current? 

Solution. Tan (p = ^ = 1.666, whence from a table of natural 
3 
tangents or logarithmic functions we find ^ = 59° 2' 6'\ 

Example. A circuit has a resistance of 2.3 ohms and an 
inductance of 0.0034 henry. An alternating current with a 
frequency of 125 passes through it. Calculate the lag. 

Solution. The reactance is 6.2832 X 125 X 0.0034 = 2.67. 
2.67 -^ 2.3 = 1. 16. This is the tangent of the angle of lag 
corresponding to the angle 49° 14' 9". 

Deduction of Ohmic Value of Capacity Reactance. If 
an alternating circuit is opened and has no capacity, no current 
can be produced in it. If capacity is present, then an alter- 
nating current will be produced by alternating e.m.f. The 
action of the capacity referred to the current wave is the following : 
As the wave starts from zero value and rises to its maximum 
value, the current is due to the discharge of the capacity, which 
would be represented by a condenser. In the case of a sine 
current the period required for the current to pass from zero 
value to maximum value is one-quarter of a cycle. At the 
beginning of the cycle the condenser is charged to the maximum 
amount it receives in the operation of the circuit. At the end 
of the quarter cycle, when the current is of maximum value, 
the condenser is completely discharged. The condenser now 
begins to receive a charge, and continues to receive it during 
the next quarter of a cycle, its charge attaining its maximum 
value when the current is of zero intensity. 

It follows from the above that the maximum charge of a 
condenser in an alternating circuit is equal to the average value 
of the current multiplied by the time of charge, which is one- 
quarter of a cycle. If practical units are used the result will be 



234 ELEMENTARY ELECTRICAL CALCULATIONS 

given in coulombs. As the period of a cycle is the quotient of i 

divided by the frequency, the quarter of a cycle is — > and the 

4/ 

value of the charge at the end of the quarter cycle is I^^ X — v 

4/ 
The e.m.f. of a condenser is equal to the quotient of the charge 

divided by the capacity. Calling the capacity of a circuit K we 

have as the value of the e.m.f. due to the capacity 



^-""vh 



2 2 imax 



AfK 



But 7^^= 7jjj^^ X -> or ;and substituting this value of I^^ 

n Tz 

in last expression gives as the value of the e.m.f. due to capacity 
at the point of maximum value, which e.m.f. is opposed to the 
impressed e.m.f. and therefore is counter e.m.f., 

-1 max 



Maximum counter e.m.f. of capacity = 



2 7zfK 



By Ohm's law e.m.f. =K/, therefore as -~|^ = /^ax X ^ 



2 7r/Z "^^-^'^TzJK 
it follows that — -rzz is the ohmic equivalent of the capacity 

2TZJK 

reactance, or virtually expresses the resistance equivalent of 
capacity. This gives the formula 

Reactance of capacity 



2 7lfK 



Example. If a 35-microfard capacity is introduced in a 
circuit of 125 frequency, what will its reactance be? 
Solution. Substituting in the formula we find 

Reactance = :: = 36.4 ohms. 

2 ttX 125X0.000,035 

Combined Impedance of Inductance and of Capacity. 

It will be seen from what has been explained in the last few 



ALTERNATING CURRENT 235 

lines that the two reactances work in opposition to each other 
in the sense that the reactance of induction acts in direct pro- 
portion to the quantity 2 nfL and the reactance of capacity in 
inverse proportion to the quantity 2 TcfK. The net reactance 
due to both, when both are present in a circuit, is obtained by 
subtracting one from the other. 

Example. A current has a frequency of 150. It passes 
through a circuit of 22 microfarads capacity and of 0.015 henry 
(15 millihenrys) inductance. Calculate the reactance of the two. 

Solution. The inductance reactance is 2 ttX 150 X 0.015 = 
14.14 ohms. 

The capacity reactance is — = 48.2^ ohms. 

2 ;rX 150X0.000,022 

The total reactance of the circuit is 48.23 — 14.14 = 34.09 
ohms. 

Lead of Current. Capacity acts as regards lead and lag in 
the reverse sense of inductance; it causes a lead of the current, 
the tangent of the angle of lead being given by the quotient of 
its reactance divided by the resistance of the circuit. The 
tangent is given a negative sign because lead is opposed to lag 
and because the positive value is assigned to lag. The formula is 

I 
reactance 2 nfK 



Tangent of angle of lead = tan ^ = — . 

° resistance R 

Lag or Lead due to both Reactances. Where both induc- 
tance and capacity are present the tangent of the angle of lag 
or of lead as the case may be is the algebraic sum of the two 
reactances divided by resistance. If the sign is positive it is an 
angle of lag; if the sign is negative it is an angle of lead. 

The giving of the positive and negative signs as described and 
the application of algebraic addition distinguishes between lead 
and lag. Otherwise subtraction can be used if attention is given 
to whether lead or lag preponderates. 



236 ELEMENTARY ELECTRICAL CALCULATIONS 

Impedance of Inductance, Capacity, and Resistance 
combined. When the three qualities of inductance, capacity, 
and resistance are present in a circuit the impedance is equal to 
the square root of the sum of the resistance squared plus the 
reactance squared. The reactance in this case is the algebraic 
sum of the two reactances, as just described. 

The equation for impedance when inductance, capacity, and 
resistance are present in a circuit is 

Impedance = Kf R^ + [2 TtfL —J . 

If reactance is due to capacity alone it has a negative sign, 
and if due to both inductance and reactance the sign will be 
negative if the capacity reactance is larger than the inductance 
reactance. But as the reactance is squared in the expression of 
resistance and as the square of a negative quantity has a positive 
sign, both reactances in all cases go to increase impedance. 

Example. Calculate the impedance of a circuit carrying a 
150 frequency current, with a circuit resistance of 23 ohms, 
inductance of 0.041 henry, and capacity of 51 microfarads. 

Solution. The reactance of inductance is 27z X 150 X 0.041 
= 38.64 ohms. The reactance of capacity is 
I 



2 t: X 150 X 0.000,051 



= 20.8 ohms. 



The impedance is \/{27,Y + (38.64 — 20.8)^ = 29.11 ohms. 

Angle of Lead or Lag. The tangent of the angle of lead 
or of lag is equal to the algebraic sum of the reactances divided 
by the resistance. The reactance of capacity is to be given a 
negative sign. . The equation is 



2 7:fL- 



TzfK 



Tangent of angle of lag or of lead = tan </> = 

K 

If the tangent is negative the angle is an angle of lead ; if 
positive it is an angle of lag. 



H 



ALTERNATING CURRENT 237 

Example. Calculate the angle of lag or lead in a circuit of 
18 ohms resistance, 0.027 henry and 47 microfarads, with a 
current of 133 frequency. 

Solution. We have 



2 ;: X 133 X 0.027 — 



Xan<}5)= 271X133X0.000,047 

18 

22.i;6 — 2^.46 , 

= ^ o = — O.161. 

18 

From a table of natural tangents we find ^ = 9° 9', the angle 
of lead, because its tangent has a negative sign. 
Proof of the Law of the Angle of Lag and Lead. — If an 

alternating e.m.f. is impressed upon a circuit, an alternating 
current will be produced whose frequency and form will be 
those of the impressed e.m.f. 

At any given instant of time the value of e.m.f. required to 
produce the current existing at that instant will depend upon 
the resistance, inductance, and capacity of the circuit. The 
algebraic sum of the ohmic equivalent of inductance and the 
ohmic equivalent of capacity, taking the latter as of negative 
sign, is the ohmic value of reactance. It follows that the value 
of the current due to a given e.m.f. is determined by the two 
elements, resistance and reactance. 

From Ohm's law we have E = RI. The value of / or of 
current in a half cycle varies from a maximum /max to zero. 
In the absence of reactance the e.m.f. required to produce 
the current existing at any instant is equal to the product of 
the resistance of the circuit by such instantaneous value of the 
current or i?/inst- 

As reactance is due to variation in strength of current, it 
follows that at any point where the current is of constant value 
there is no reactance, In an alternating-current wave such 



238 ELEMENTARY ELECTRICAL CALCULATIONS 



point coincides with the maximum current value, which is at 
the summit of the sine curve. Here the tangent to such curve 
is parallel to the base line and the current for an instant is of 
constant value, which value is, of course, /max- The e.m.f. 
required to produce such current depends entirely on the 
resistance of the circuit, and is expressed by RItoso.- This 
value is the height of the ordinate of the e.m.f. curve at the 
point where the current is of maximum value. As the e.m.f. 
curve is a sine curve, the value of the ordinate in terms of Ejnax. is 
^max sin d. 




The rate of change in an alternating current is greatest where 
its sine curve crosses the base line, at which point the current is 
zero value, and which point is 90 degrees removed from the 
summit of the curve. 




The value of the impressed e.m.f. at this point depends 
entirely upon the reactance of the circuit. Resistance does not 
affect the value, because the current is of zero value. CalHng 
the reactance ^, we know that if we multiply it by the maximum 
value of current we have the counter e.m.f. due to reactance. 
Therefore the e.m.f. required to produce an alternating current 



ALTERNATING CURRENT 239 

at the point of zero value of current where resistance plays 
absolutely no part is equal to xlmas.- The expressions RImax. 
and xlmax are the values of the two ordinates of the sine curve 
of the impressed e.m.f. They are proportional to sines of 
angles, which angles differ in value by 90 degrees; in other 
words, they are proportional to the sines of complementary 
angles, and by trigonometry they are the sine and cosine respec- 
tively of such angles to the radius £max- In terms of £max these 
expressions become i?/max = -Emax sin 6 and ^/max = ^max sin 
(d + 90°) = £max cos d. 

To obtain the absolute value of one of the angles, the value of 
the ordinate erected at such angle is divided by the value of the 
ordinate erected on the other angle; the result by trigonomtery 
is the tangent of the angle in question. The angle d now 
becomes the angle of lead or lag; if we call it (f) we have 



m^ - ^ ^ -Emax sin ^ ^^^ ^^ 



xlj 

jR^max R -Emax COS (f) 



or reactance divided by resistance gives the tangent of the angle 
of lag or of lead. If the ohmic value of the capacity reactance 
exceeds that of the inductance reactance, giving a negative 
sign to tan^, the angle is an angle of lead and vice versa. 

Power and Power Factor. The power of an alternating 
system is equal to the product of the effective e.m.f. and effective 
current, provided there is neither lag nor lead. In this case the 
product of the e.m.f. and current value is always the product 
of two negative or of two positive quantities ; hence it always has a 
positive sign. But when there is lag or lead there is sometimes 
a product of positive e.m.f. by negative current, sometimes of 
negative e.m.f. by positive current; both of these products are 
negative. Then there are products of current by e.m.f. where 
both are of the same sign, which products are positive. The 
algebraic sum of the products for all angular values gives the 



240 ELEMENTARY ELECTRICAL CALCULATIONS 

power, which in the case of a lag is evidently less than when there 
is no lag. The power of an alternating current system is equal to 
the product of the effective current by the effective e.m.f. multi- 
plied by the cosine of the angle of lag. The cosine of the angle 
of lag is called the power factor. 

Example. Assume an angle of lag of 49° 14' 9'' and calcu- 
late the power, the effective current being 34 amperes and the 
effective e.m.f. 150 volts. 

Solution. The cosine of the angle of lag 49° 14' 9" is 0.65295. 
The product of the effective e.m.f. by the effective current by the 
power factor gives the power; it is 150 X 34 X 0.65295 = 
3,330 watts. If there were no lag or lead the power would be 
150 X 34 = 5,100 watts. 

Example. What is the power in a circuit with a lag of 
90 degrees ? 

Solution. As the cosine of 90 degrees is zero, the power of such 
a circuit is zero. 

The power factor is thus deduced by trigonometry : 

Let the angle of period or of position of an alternating e.m.f. 
be designated by d and the angle of lag of the current by ^. 
Then the angle of period of the current will he d — <f>. Then the 
values of e.m.f. and of current will be at the time of this period 

E=E^,^smd (i) 

and / = /j^ax sin (d - ^). (2) 

If these equations are multiplied together the products will 
be the power of the circuity at that time. This gives 

EI= £„,, /„,, sin sin (d - <j>). (3) 

By trigonometry sin (d — (j)) = sin d cos ^ — sin (^ cos 6, and 
substituting in (3) gives 

£^ = ^max ^max sin'^ COS (f) - E^^^ I^^^ siu COS d sin (f). (4) 

The angle cf) is invariable, but for function? of d the average 



k 



ALTERNATING CURRENT 24 1 

values must be substituted. The average value of sin^^ is i, in 
accordance with what has been proved on page 227. The 
average value of sin (9 cos ^ is o, because it has an equal sum of 
positive and of negative values whose average is o. Substituting 
these values in equation (4) we have 

Average EI = ^°^^^ ^"°"^ cos ^, (5) 

the last term reducing to o because one of its factors, sin 6 cos d, is 
equal to o. 

F T 

We have found that E^f = -^and /ef = -^ ' and substitut- 

V2 V2 

ing these values in equation (5) we have 

Average power (average EI) = E^^ I^^ cos <^. (6) 

Power Factor for both Reactances Combined. The power 
factor applies to capacity reactance exactly as it does to induc- 
tance reaction, and consequently applies also to the combina- 
tion of the two reactances. From the practical standpoint the 
angles of lag and of lead are always treated as if they lay in the 
first quadrant of the circle. Even the negative sign of the tan- 
gent (j) when it occurs is simply used to determine whether the 
angle is one of lag or of lead, but in finding the value of the angle 
from a table it is treated as a positive quantity. The power 
factor, which is a cosine of an angle between o degrees and 
90 degrees, must always have a positive sign. 

Example. What is the power factor in the circuit of the first 
problem on page 237 at the given frequency? 

Solution. It is the cosine of 9° 9' =0.987. 

If the inductance in henrys of a circuit is numerically equal 
to the capacity of the same in farads, there will be no reactance, 

and the tangent of ^ will be expressed by — =0, because the 



242 ELEMENTARY ELECTRICAL CALCULATIONS 

reactance, which is the numerator of the expression for the 
tangent of (j), is of o value. The angle whose tangent is o is 
the angle of o degrees; hence when there is no reactance there 
is neither lag nor lead. The cosine of o degrees is i.ooo; hence 
when there is no angle of lag or of lead the power factor is i, 
and the power is given by the product of the effective e.m.f. 
by the effective current. 

Angle of Lag and Rate of Change. The maximum rate of 
change of current value occurs when the value of cos 6 is greatest, 
and this is when the sine of the curve is of zero value or when 
the sine function, in this case the current, is of zero value. By 
the law of inductance the product of the rate of change by the 
inductance of the circuit gives the counter e.m.f. This counter 
e.m.f. must be equal and opposed to the impressed e.m.f.; in 
other words the point of zero value of current must correspond 
with the point where the impressed e.m.f. has the value of the 
maximum counter e.m.f. of inductance and is opposite to it in 
polarity. If the current wave or cycle is taken as shifted back- 
wards to bring about the equal opposed relation described, the 
angular distance through which it is shifted is called the angle 
of lag, and the maximum counter e.m.f. is evidently the sine of 
the angle of lag. 

When two parts of an alternating circuit are in parallel with 
each other the combined impedance of the two branches is cal- 
culated in a way similar to that employed for the case of direct- 
current parallel circuits. 

If the impedances on both branches are due to the one cause, 
i.e., to resistance, to capacity, or to inductance, the combined 
impedance is equal to the reciprocal of the sum of the reciprocals 
of the impedances exactly as in the case of parallel circuits 
carrying direct currents. 

Example. Two branches of a circuit are in parallel, each being 
of negligible resistance and inductance. The capacity of one is 



ALTERNATING CURRENT 243 

TO microfarads, the capacity of the other is 5 microfarads ; the 
frequency is /. Calculate the combined reactance. 

Solution. The reactances of the branches are respectively 

;: and ;: ; the sum of their re- 

2 nf X 0.000,010 2 TT/ X 0.000,005 

ciprocals is 2 nfX 0.000,015, and the reciprocal of the sum of 

their reciprocals is ■ . This is simply the re- 

2 TT/ X 0.000,015 

actance of a capacity of the sum of the two capacities in parallel. 

Placing capacities in parallel with each other is equivalent 
to the production of a capacity equal to the sum of the capacities 
in parallel with each other. 

Example. Two inductances each of 0.03 henry are in 
parallel on a 125 frequency system. What is their combined 
reactance, assuming the resistances and capacities to be negli- 
gible in amount ? 

Solution. The reactance of a single inductance is 2 nfL, or 

2 ;r X 125 X 0.03 = 23.562 ohms equivalent. Both are of the 

same value by the conditions of the problem; the sum of their 

1 . I I 2 

reciprocals is H — = — > the reciprocal of which 

23.562 23.562 23.562 ^ 

2 ■? C62 
is -^^ — =11.781 ohmic equivalent. 
2 

The reactance of two equal inductances in parallel is one-half 
their sum. 

Example. Let the two inductances be of 0.05 and 0.07 henry, 
the other conditions remaining the same as in the last problem. 
Calculate the combined reactance. 

Solution. The reactance of the inductance 0.05 is2 7rX 125X 
0.05 =39.27 ohmic equivalent; that of the inductance 0.07 is 
2 ;r X 125 X 0.07 = 54.978. The sum of their reciprocals is 

and the reciprocal of this sum of 



4. I _ 94.2 48 



39.27 54.978 2,158.986 

the reciprocals is 22.9075 ohmic equivalent. 



244 ELEMENTARY ELECTRICAL CALCULATIONS 

This problem is conveniently done by logarithms, thus : 

log 39.270 1.594,061 39-270 + 44-978 = 94.248 ' 

log 54.978 1-740,189 

log of product 3.334,250 log of sum 94.248 as above 1.974,272 

log of product 3 .334,250 

log of sum 1 .974,272 

log Qf ^^sum^ 1-359,978 

Ohmic equivalent = 22.9075. 

Such problems can be done approximately with the slide rule. 

For resistances in parallel in an alternating-current circuit the 
calculation is the same as for direct current. 

Let an inductance and a resistance be in parallel with each 
other, and let an alternating e.m.f. be impressed upon them. 
It will be the same in all respects for both. Its instantaneous 
maximum value in terms of the inductance reactance will be 
^max =^'max 2 tt/L (i), and in terms of the resistance of the 
other branch will be ^^^ax ^ ^"vi2.^ -'^ (2), ^' indicating the current 
through the inductance and I" the current through the resistance. 
The instantaneous values of the two currents are in quadrature 
with each other, so the resultant current is given by the expres- 
sion /^ax = V^^^max + ^'''max (s)- From (i) and (2) we 
have V^,^ = f^and /"™ax = %-"' and substituting in (3) 
these values gives 



J _ \ / °^^^ J- °"^^ 7? \{ t / -*• 1_ J (a\ 

™^^ V (2 TzfLf "^ i?2 -^max ^ V (2 ^j^2 -t- ^2 i4; 

From Ohm's law /=£X^, so that it follows that 



si 



( Ttv' "*" "pT ^^ ^^^ reciprocal of the ohmic equivalent 
of the impedance of the parallel inductance and resistance, 



ALTERNATING CURRENT 245 

and the ohmic equivalent of the impedance of the two branches 
is the reciprocal of the square root of the sum of the recip- 
rocals of the squares of the resistance and reactance. 

The same law applies to the reactance of capacity and is 
demonstrated in exactly the same way. 

Example. Calculate the impedance of a resistance of 30 ohms 
in parallel with an inductance of 0.03 henry. The frequency is 
100 cycles. 

Solution. The inductance reactance is 2 ;r X 100 X 0.03 = 
18.85 • Reciprocal of impedance = 



^- 



H r = v 0.003,925 = 0.06265 . Impedance^ 



18.852 302 —o.y-o o r 0.06265 

= 15.96 ohmic equivalent. 

If capacity and resistance are in parallel, the impedance is 
equal to the reciprocal of the square root of the sum of the 
reciprocals of the squares of the resistance and capacity 
reactance. 

Example. Calculate the impedance of a resistance of 20 
ohms in parallel with a capacity of 90 microfarads with a 
frequency of 125. 

Solution. The capacity reactance is = 

2 ;: X 125 X 0.000,090 

14.147. The reciprocal of the impedance is 

Vf— Y + f-^^y=i/— + —nearly =0.0816. 
▼ \20/ \i4.i47/ V 400 200 

Therefore the impedance = — — — = 12.255 the ohmic equiv- 

0.0816 

alent. 

PROBLEMS. 

What is the length of a sine curve ordinate at a point 0.8 radian 
from the origin or commencement of the base line? Take the value 
of a radian as 57.3°. Ans. 0.717. 

If the base line is 3 inches long, where should the above ordinate be 
placed and how long should it be ? ■ In this case 27t is represented by 



246 ELEMENTARY ELECTRICAL CALCULATIONS 

a length of 3 inches, and the radius of the generating circle in inch 
measurements is 3/2;:. 

Ans. At 0.38 in. from origin; 0.34 in. high. 

If the maximum e.m.f. of the above cycle is 2,250 volts, what is the 
voltage at the above point or period ? Ans. 1,614 volts. 

Calculate the average value of the e.m.f. of an alternating circuit of 
1,275 volts maximum e.m.f. -^^^^ 811.665 volts. 

If the average value of an alternating current is 97 amperes, what is 
its maximum value? Ans. 152.29 amperes. 

What is the effective value of an alternating e.m.f. whose maximum 
value is 4,500 volts? Ans. 3,181.5 volts. 

The effective value of an alternating current being 97.7 amperes, 
what is its maximum value? Ans. 138.15 amperes. 

Calculate the reactance on a circuit due to an inductance of 0.172 
henry and a frequency of 175. Ans. 189.12. 

An alternating current requires 1/250 second for a cycle. The 
inductance of the circuit is 0.091 henry. Calculate the reactance of 
inductance. Ans. 142.94. 

With an angular velocity due to 125 alternations per second and an 
inductance of 0.078 henry, apply the angular velocity formula to the 
calculation of the reactance of inductance in a circuit with the above 
frequency. Ans. w = 785.4 radians; reactance = 61.26. 

Calculate the rate of change of a 29-ampere (effective), 55 frequency 
current at 60°, Ans. 7,084 amperes. 

Calculate the impedance due to a resistance of 179 ohms and an 
inductance of o.ii henry with a frequency of 125. Ans. 199. 

Calculate the impedance of a resistance of 251 ohms and an induc- 
tance of 0.901 henry with a frequency of 75. Ans. 493. 

What is the angle of lag or of lead in the above case ? 

Ans. tan = 1.6916; angle of lag = 59° 24'. 

With a resistance of 14 ohms, an inductance of 0.09 henry, a capacity 
of 31 microfarads, and a frequency of 75, calculate the impedance of a 
circuit and the angle of lag or of lead. 

Ans. Impedance 29.56; tan = — 1.8601; angle of lead = 61° 44'. 

In the last example let the maximum e.m.f. be i,25ovolts. Calculate 
the effective e.m.f. and current and the power factor and the power. 

Ans. £eff. = 883.75 volts; IqQ^= ^2g.8g amperes; power factor = 
0.47358; power = 12,510 watts = 16.77 horse-power. 



CHAPTER XVIII. 



NETWORKS. 

Cycles and Meshes. — Direction of Real Current. — Outer and Inner 
Meshes. — Indication of Cycles. — Cycle Equations. — Cycle Calcu- 
lations. — Meaning of Cycle Letter. — Currents in a Network. — 
Maxwell's Rule for Writing the Equations of the Cycles. — Resist- 
ance of a Network. — Fleming's Method of Calculating the Resist- 
ance of a Network. 

Cycles and Meshes. Any quantity can be expressed as the 
difference of two other quantities both of the same or of different 
signs. Thus i ampere can be expressed as the difference of lo 
and 9 amperes or as 10-9 amperes. Instead of numbers letters 
can be used whose difference, sls x — y, is taken as numerically 
equal to the real quantity. Thus the real quantity i ampere can 
be represented by x — y ii the condition is imposed on the 
unknown quantities that they shall differ by unity, or that x — y 
shall equal i. 

Assume two meshes of electric conductors such as indicated 
in the cut, Fig. i. Let them be connected at C and D to a, 
source of e.m.f. Their relative resist- 
ances and connections may be such 
that a current will pass through AB or 
that none will pass. Assume in each 
mesh imaginary currents x and y cir- 
culating in the direction opposed to 
that of the hands of a watch and in- 
dicated by the curved arrows. Call this direction positive. 
The intensity of the current through AB will be represented 
by. the difference of the imaginary currents x and y. Then 
x — y = the current in AB. Such imaginary currents are 
called " cycles.'* 

U X = y, the real current in AB will be of zero intensity, 

247 





x=A 




248 ELEMENTARY ELECTRICAL CALCULATIONS 

as in the balanced Wheatstone bridge. If :x; > y, there will be 
a current equal to x — y from B to A. li x < y^ there will 
be a current equal to y — x from ^ to -S. 

The subtraction is to be made algebraically, giving each cycle 
its sign, positive or negative as the case may be. Thus in Fig. 2 
^ let :x: = I and y = — 2. Then y — x = ~ 2 

— I = — 3, or the real current has a value 
of 3. On referring to Fig. 2, which repre- 
sents this case, it is evident that the current 
in a lead such as AB lying between two cycles 
g of opposite sign should have a value equal 

^8-2. to their sum. 

Direction of Real Current. The direction of the real cur- 
rent can be referred to the adjacent portions of one or both of the 
cycles. In Fig. 2 the current is from ^ to ^, corresponding to 
that of both cycles. This is a case where the two cycles are of 
opposite sign. When both cycles have the same sign the direc- 
tion of the real current corresponds to that of the numerically 
larger cycle. In Fig. i, if a; = i and y = 2, then the direction of 
the real current in AB corresponds to that of the adjacent side 
of y, or is from A to B. 

The algebraic rule for direction follows from what has been 
said. Subtract algebraically one of the cycles from the other 
and refer the direction of the real current, whose value is thus 
found, tp the minuend. If the real current has the same sign as 
the minuend its direction is the same as that of the adjacent part 
of the minuend. If of different sign from the minuend its 
direction is opposite. 

Suppose we have two adjoining cycles x = $ and y = — S- 
Taking x as the minuend we have 

x-y = s + S=^^' 
The value of 10 is positive, as is also the minuend ; x gives the 



NETWORKS 



249 



direction of the real current. Taking y as the minuend we have 



10, 



y- ^ = - 5 - 5 = 

which tells us that the direction is the same as that of x. On 
looking at Fig. 2, which represents this condition, we see that the 
adjacent sides of both x and y coincide in direction, and both 
results are true and identical. 

The meshes of Fig. i are shown without any source of e.m.f. 
Such source, a battery for instance, might be placed in any mem- 
ber. Or an outer source of e.m.f. may be introduced. If such is 
introduced it must inevitably introduce another mesh, as shown 
in Fig. 3. It is impossible to connect an outer source of e.m.f. to 
a mesh or meshes without thereby adding one more mesh to them. 

In the diagram of the battery at £, Fig. 3, let the small bar 
represent the carbon or copper pole. Then the e.m.f. is of such 
polarity that it tends to produce a current from E to C, C to D, 
and D to E. The cycle x expresses this direction as drawn, and 
has a negative value because its direction is that of the movement 
of the hands of a watch. To the e.m.f. of a battery tending to 





produce a negative cycle is to be assigned a negative sign. If this 
battery has an e.m.f. of 1.75 volts, it is written — 1.75 for cycle 
calculations. The e.m.f. of a battery tending to produce a 
positive cycle has a positive sign. 
If the battery mesh is supposed to be swung or rotated up- 




250 ELEMENTARY ELECTRICAL CALCULATIONS 

ward as in Fig. 4 no change will occur in the relations of the real 

currents, but the cycles x and y become reversed and are now of 

positive sign. Therefore the battery becomes of positive e.m.f. 

and its voltage is written 1.75. 

Outer and Inner Meshes. An outer mesh is one which has 

one or more sides on the outside border of the network. All the 

meshes in the diagrams Figs, i, 3, and 

4 are outer meshes. An inner mesh is 

one all of whose sides lie within the 

border of the network. In Fig. 5 an 

inner mesh is shown at A ; the other 

four meshes are outer meshes. An inner 
Fig. 5. 

lead is a conductor forming part of two 

meshes. The lead AB in Fig. 4 is an inner lead. An outer 

lead is one lying on the border of a network. CB and BD in 

Fig. 4 are outer leads. 

No cycle can be assumed to exist in space outside of a mesh. It 
follows that the current in the outer lead of a mesh is equal to the 
value of the cycle, because there is no cycle outside the network 
and adjacent to such lead, and therefore there is nothing to be 
added to or subtracted from the mesh cycle, which cycle there- 
fore is the value of the real current. 

Referring to Figs. 3 and 4 it is evident that the values and 
directions of the real currents are unaffected by the position of the 
active loop. In Fig. 3 the current in CA is equal to z, because 
CA is an outer lead, as explained in the last paragraph. In Fig. 4 
the current in the same lead, CA^ is equal io x — z, because 
CA is now an inner lead. It follows that the values of the cycles 
are different in the two diagrams, unless all the leads of ACBD 
are of the same resistance. The value of network cycles depends 
upon the position of the active loop. This does not apply to 
the real current intensities, which are unaffected by such vari- 
ations of position. 



NETWORKS 251 

Indication of Cycles. Cycles are expressed by letters as 
symbols, like unknown quantities in algebra, as x^ y, z, etc. 
They are used as the bases of simultaneous equations, one for each 
mesh, and therefore there are as many equations as there are 
meshes or cycles. One at least of the equations has a known 
quantity (e.m.f.) in the second term. Hence by the rules of 
algebra the equations can be solved, and the values of the 
unknown quantities, which are the imaginary currents in the 
cycles, are determined. By subtracting the cycle values from 
each other for inner leads, or by taking them unreduced for outer 
leads, the values of the real currents in the most intricate networks 
can be determined directly. 

The signs of the cycles come out in the operation as 
+ or — signs prefixed to the numerical values of the cycles as 
determined by means of the simultaneous equations. Suppose 
cycle X and cycle y lie next to each other. Their value is cal- 
culated. Assume that x proves to have a value of 5 and y to 
have a value of —5. Then we find the value of the current 
through the lead lying between x and y by simply adding the 
values of x and y and finding its direction by inspection of the 
diagram or algebraically by the signs. 

The cycles x, y, etc., are all taken as of positive sign in the 
original equations. The known quantity or quantities to which 
the first member or members of one or more of the equations 
are equated are the voltages of the sources of e.m.f. acting on 
or taken as acting on the system and lying directly on one of 
the sides of a mesh, so that the mesh, if all other meshes were 
removed, would, with the generator, be an active electric cir- 
cuit. This quantity is taken as positive if the polarity of the 
generator is such that it would tend to produce in its own 
mesh a cycle opposed in direction to the watch hands. If it 
would tend to produce a cycle of the same direction as the 
watch hands it is given a negative sign. 



252 ELEMENTARY ELECTRICAL CALCULATIONS 



Cycle Equations. In writing a cycle equation the mesh 
inclosing the cycle is treated as a closed electric circuit. If a 
generator of e.m.f. is included in the circuit of the mesh, the 
first member of the equation of the cycle is equated to the value 
of the e.m.f. of that generator. If there is no generator in the 
circuit of the mesh, the first member of the equation of the cycle 
is equated to zero. Suppose that there are n meshes in a net- 
work and that there is a generator on an outer lead of an outer 
mesh. Then the cycle equation of that outer mesh will be 
equated to the e.m.f. of the generator and the remaining n —i 
equations will be equated to zero. Suppose that the generator 
lies between two meshes of a network. Then the cycle equa- 
tion of each of these meshes will be equated to the e.m.f. of the 
generator. For one equation the e.m.f. will be positive, for the 
other it will be negative, and there will he n — 2 equations 
equated to zero. In all cases the sign to be prefixed to the 
e.m.f. of the generator is to be determined as by the consider- 
ations just given. An example of cycle calculations will illus- 
trate what has been said above, and will show the significance 
of cycle values and of cycle signs. 

^, Cycle Calculations. The current in any 

lead will be indicated by its terminal let- 
ters large size, the resistance of any lead by 

B| f IC its terminal letters small size. Thus in the 

lead AB, Fig. 6,, the current would be in- 
dicated by the same letters, AB^ the resist- 
I ance by ab. 

Assume the network indicated in the dia- 

L '^ gJ'am Fig. 6. The battery at V is the 

Fig. 6. source of e.m.f. As it tends to produce a 

cycle current of the opposite direction to that of the motion of 

the hands of a watch its e.m.f. has a positive sign given to it. 

Assume that every member of the network has the same resist- 







00 


® 


3 









NETWORKS 253 

ance, ab = be — eg^ and so on. We know from Ohm's and 
Kirchhoff's laws the general distribution of current through the 
members of the network. 

No current will go through AL because the circuit is balanced. 
This condition is expressed by stating that the left-hand cycles 
are equal respectively to the right-hand ones in pairs. Thus 
x=oc^,y=y\z= z' . 

The lead BE is an outer lead; the real current in it is equal 
to x^ or BE = X. The same considerations tell us that EG = y 
and GX = z. We know from Kurchhoff's law that BE > EG 
and that EG > GK. Therefore x > y djid y > z. The real 
current through ED \?> equal to ;x; — ;y. Asy x > y and as x is 
positive a current passes through ED from E to D^ this direc- 
tion being determined by the cycle of the minuend. This is 
in exact accord with Kirchhoff's laws, and the same process can 
be applied to other meshes, and results in accord with the known 
laws of the distribution of current will be obtained in every case. 

The leads from the battery V io B and C are outer leads; 
therefore the current in them is equal to the cycle current u. By 
Kirchhoff's law u or VB = BA -f BE. Applying cycles we have 
u — X =^ AB, 
x=BE. 
Adding these we have ^ _ ^^ , «^ 

exactly in accordance with Kirchhoff's law. 

In like manner the current x or BE divides itself between 
ED and DG by Kirchhoff's law, or 

X or BE =ED -\- EG. 
Applying cycles we have 

X- y =ED, 
y = EG. 
Adding these we have x = ED A- EG 

as before, in accordance with Kirchhoff's law. 



254 ELEMENTARY ELECTRICAL CALCULATIONS 

By appl)dng this system of examination it would be found 
that by the application of cycles Kirchhoff's law is rigorously 
carried out. 

Meaning of the Cycle Letter. The cycles, designated 
each by a single letter, are the basis of the determination of the 
currents in the different members of networks under given 
e.m.f. and of the determination of the total resistance of any 
network between given points. The letter designating the cycle 
means the number of amperes constituting the imaginary 
current of the cycle. Thus while it is convenient to speak of 
the cycle x, y^ or a, as the case may be, the letters x, y, or a really 
stand each for a definite number of amperes. The value in 
amperes of each cycle is determined by simple algebraic methods 
next to be described. The knowledge of these values gives the 
distribution of currents in the network. If it is only required 
to know the resistance of a network, a definite e.m.f. is assumed 
to be connected to the points between which the resistance 
is required to be known. The generator and its connections 
make an extra mesh whose resistance is taken at zero. There 
is only one e.m.f., that of the generator, and the current through 
that mesh as a divisor divided into the e.m.f. as a dividend 
gives as quotient the resistance of the network. 

Currents in a Network. To determine the currents in the 
leads of a network an equation has to be written for each mesh. 
These are called cycle equations, or equa- 
tions of the X cycle, of the y cycle, and so 
/^ ^^ on. Assume the network represented in the 




® © 



diagram Fig. 7. Let the potentials at the 
points of intersection be designated by the 
letters at such points. Then B — C would 
p.g^ represent the fall in potential in the entire 

network below BC and including BC. By 
Ohm's law this is equal to the e.m.f. of the battery or gen- 



D 



NETWORKS 255 

erator at V less the product of the current in the battery- 
leads VB and VC by the resistance of the battery and same 
leads. This resistance we will call v. It is the resistance 
from 5 to F to C, including leads and generator. The e.m.f. 
of the generator we will call e. The current in the leads 
VB and VC^ which are outer leads, is equal to u. Then 
by Ohm's law, as stated above, we have 

e — vu = B — C. 

In this equation u indicates the current of the battery mesh. 

The fall of potential in BA is equal io B — A. It is also 
equal to the product of the current in BA^ which is u — x^ by 
its resistance, which we have decided to term ha. 

(u- x){ba) = B - A. 
By exactly similar process we find 

(u - x") (ac) =A - C. 
Transposing each equation we obtain 
e = B — C + vu. 
o = A — B + (ba) (u — x). 
o = C — A -\- (ac) {u — x'). 

Adding these together we obtain 

e = vu •\- ba {u — x) -Y ac {u — x/). 
Transposing and grouping this we find 

u {v -{■ ha -\- ac) — x (ba) — x' (ac) = e. 

This is the equation of the cycle u. 

The same result may be obtained in a shorter way. In the 
circuit VBACV the currents are divisible into three; a current 
through CVB which is equal to u; one through BA which is 
equal to u — x; and one through AC which is equal to 
u — x^. By Ohm's law the sum of the e.m.f. 's in the circuit 
is equal to the product of resistances by currents, each 



256 ELEMENTARY ELECTRICAL CALCULATIONS 

resistance being multiplied by the current passing through 
it. The e.m.f. in this circuit is e. Call the resistance of the 
generator and of its leads YB and YC simply v^ and for the 
other resistances use the terminal letters of the respective leads, 
small size. We have therefore 

vu + (6a) {u -\- x) •\- (ac) (u — x') = e. 

Grouping this we have as before 

u (v + ba -}- ac) — X {ha) — x' {ac) — e. 

The equation for a cycle in a mesh which contains no source 
of e.m.f. in its circuit may next be deduced. Take the mesh 
ABED J whose cycle is designated as x. Assume a source of 
e.m.f. at any point, say at B, and let its value in volts be desig- 
nated by e\ As the leads BE and ED are outer leads the 
current through them is equal to x. The current x multiplied 
by the resistances he and ed will, by Ohm's law, give the e.m.f. 
expended on BE SindED. If this is subtracted from the total 
e.m.f. of the circuit it will, by Kirchhoff's first law, give the e.m.f. 
expended on DAB, which, if we denote the potentials at the 
corners by letters placed there, is equal to D—B. This gives 
e - X {he -{■ ed) = D - B. 
X — x^ is the current in A D. Multiplied by the resistance 
ad of the lead AD it gives the e.m.f. expended on that lead, and 
the parallel operation holds for AB giving 
{x — x^) ad = D — A, 
{x — u) ab = A — B. 
Transposing gives 

{be+ ed) x + D - B = e, 
(ad) {x- x') +A -D =o, 
(ah) (x- u) + B - A =o. 
Adding and grouping we have 

X {be + ed + ad + ab) —x' {ad) — u (ah) =e'=o. 



NETWORKS 2-57 

This is put equal to zero because there is no e.m.f. in the 
mesh. 

This method may be applied as follows : 

In the mesh BE DA, multiply the resistance of each lead by 
the current flowing through it, which gives, by Ohm's law, the 
e.m.f. expended on each of the leads. Added together they give 
the e.m.f. expended all around the circuit, which e.m.f. is zero, 
because there is no generator in the circuit. The equation thus 
obtained is 

(be) X + (ed) x + (ad) (x — x') + {ah) {x — u) = o. 

Grouping this gives the same equation found above. 

X {he + ed -\- ad + db) — x^ (ad) — u (ab) — o. 

We have now found cycle equations for a mesh with a genera- 
tor in its circuit and for a mesh without one. Both are of iden- 
tical form, and the rule for writing cycle equations is apparent. 
It is this: 

Maxwell's Rule for Writing the Equations of the Cycles. 
Multiply the cyclic symbol by the sum of the resistances of the 
sides of its mesh ; subtract from the re- 
sult the products of the symbols of each 
of the adjoining cycles by the resistance 
common to it and to the mesh whose 
cyclic equation is to be written. 
Equate the result to the e.m.f. in the 

circuit of the cycle. Give to such e.m.f. 'e.?n./.= 2 

R=7 
a sign in accordance with the explana- 
tions given. 

By applying this rule a cycle equation is written out for each 
mesh, and they are solved by regular alegbraic methods. 

Example. Assume the network indicated in Fig. 8 in 
which the numbers indicate the resistances of the respective 




258 ELEMENTARY ELECTRICAL CALCULATIONS 

leads. The numbers serve also to designate the leads. The 
resistance of the battery and its leads to A and 5 is 7 ohms. 
Calculate the current in the battery lead and in the leads 8 
and 4. 

Solution. Applying Maxwell's rule, we write out by inspection 
the three simultaneous equations of the cycles of the meshes 
Xf y, and 2. 

x{t -\- S + 2)-Sy -22 =2.(1) 

-8:x; + 3'(4+7 + 8) - 7z =0.(2) 

— 2 X — 7 y + 2 (i + 7 + 2) = o. (3) 

In writing out these equations the same order is preserved for 
the three cycle symbols. Performing the indicated additions we 
have 

I'jx— Sy— 2Z = 2. (4) 

-Sx+igy-'jz=o. (5) 

— 2X— 'jy+ 10 z=o. (6) 

Equations i and 4 are the equations of the cycle x; 2 and 
5 the equations of the cycle y ; and 3 and 6 the equations of the 
cycle 2. The e.m.f. of the battery is positive because as placed 
the battery tends to produce a positive cycle as already explained. 

Solving the above in the ordinary way or by determinants we 
find 

X = 0.1936 ampere, 

y =0.129 ampere. 

As the battery leads are outer leads the current in them is 
equal to the cycle of their mesh, or to x. It is 0.1936 ampere. 

The current in lead 4 being an outer lead is equal to the cycle 
of its mesh, or y. It is 0.129 ampere. 

The current in lead 8 is equal to x — y^ or 0.1936 — 0.129 ~ 
0.0646 ampere. 



« 




NETWORKS 259 

Example. Assume the network shown in the diagram 
Fig. 9. It comprises two sources of e.m.f. 
which are so connected that one of them 
tends to produce a negative cycle and the 
other one a positive cycle. Each has 2 
volts e.m.f., and a positive sign is given 
to the battery E and a negative sign to 
the battery E\ Calculate the currents in 
the leads AEB and AE^B. The resistance 
of each battery with its leads is taken as 2 
ohms, and the resistance of AB as i ohm. 

Solution. Writing out Maxwell's formulas by simple inspection 
of the diagram we have 

X(2 + l) — 3/ = 2. 

y {2 ■\- 1) — X = — 2. 

Transposing and performing the indicated additions we 
have 

3^ - >'= 2, 

- x-v 2>y = - 2, 
which solved give 

X = — and y = — — . 
22 

As the battery leads are outer leads the current in each of them 
is equal to the cycle current in its mesh. This gives J ampere for 
each, and the direction is determined by the position of the 
battery in the circuit of the mesh. The current in the inner lead 
AB is equal to the difference of x and y. 



-y= ,or -+- 

2 \ 2 / 2 2 



- + - = I. 
2 



26o ELEMENTARY ELECTRICAL CALCULATIONS 

Example. Assume the network of Fig. lo which, after what 
has been said, is self-explanatory. Calculate the currents. 

Solution. Writing out Maxwell's equations directly and add- 
ing and grouping them directly from the figure we have 

:v (2 + 2) — 2 z = 2, 

— 2X— 2y-\-z{l + 2+l-\-2)=Of 

y (2 + 2) — 2 Z = — 2, 

which solved give 



X = — , y 
2 



, and z = o. 



Current in AB , 
Current in A'B\ 



X — z = 



o = — ampere. 
2 



1 I 

y~z = — — — 0= ampere. 

2 2 



The values of x and y give the currents in the battery leads, 
because these leads are outer ones. The value of z, which is o, 



+2 Volts 





-2 Volts 
Fig. 10. Fig ". 

gives the currents in ^^' and BB\ because they are outer leads 
also. The currents in the battery leads are ^ ohm each, and the 
currents in AA^ and BB' are of zero value. 

Example. Assume the network of Fig. 11 and calculate the 
currents. 



NETWORKS 



261 



Solution. The e.m.f. to which the equation of the cycle x is 
equated has a negative sign; the e.m.f. for the y cycle has a 
positive sign. The equations are written out by simple inspection. 
4x- y = - 2, 

- x+ 4y = 2, 

2 2 

which solved gives x = axidy = -, and the current through 

the intermediate lead AB is equal to x — y =— — _1 = _4 

5 5 5 

of the sign of the minuend and corresponding therefore in direc- 
tion with the portion of it adjacent to AB or from ^ to ^. Or 
the subtraction may be made the other way, thus : 

2 / 2\ 2 , 2 4 
y- X = =- + -=i. 

5 V 5/ 5 5 5 
The result has the sign of the minuend ; its direction is there- 
fore determined hy y. On inspection it will be seen that both 
processes give it the same direction, which is as it should be, 
and its value is | ampere. 

The above equations may be solved by determinants. To do 
this a coefficient of o must be assigned to any cycle in an equa- 
tion in which it normally would not appear, and this coefficient 
takes its place in the determinant. Thus the determinant of the 
three equations for Fig. 10 is the following: 
For the common denominator, 

4 0—2 
- 2 — 2 6 



For the numerator of x, 



2 

o 

— 2 



o 

— 2 

4 



-64. 



= - 32. 



or X 



— J- = — as above, and so on for the others, y and z. 
— 64 2 



262 ELEMENTARY ELECTRICAL CALCULATIONS 



The reason for speaking so specifically of determinants is that they 
are the basis of a very simple rule for calculating directly and by 
a single operation the resistance of any network. The method 
is described in Fleming's " Handbook for the Electrical Labora- 
tory and Testing Room," Vol. I, p. 204 et seq., and is given later 
in this chapter. 

Resistance of a Network. The resistance of a network 
varies according to the points between which the resistance is 
taken. The resistance between any two points is thus deter- 
mined : Assume that a generator of zero resistance is connected 
by leads of zero resistance to the points of the network between 
which the resistance is to be determined. Assigning any e.m.f. 
to the generator, the value of the current through the generator 
and leads is determined by cycle equations. But the only resist- 
ance through which this current has to go is that offered by the 
network. We therefore know the e.m.f. and the current; the 
resistance is calculated by Ohm's law. 

Example. What is the resistance be- 
tween A and B of the network of Fig. 12, 
the resistance of each of whose mem- 
bers is placed alongside of its center? 
Solution. Assume an e.m.f. say of 
2 volts produced by a generator at V. 
The generator and its leads are 
assumed to have no resistance. They 
introduce a third mesh with its cycle x. Writing out by 
simple inspection the cycle equations we have 
6 X — 2 y — 42 = 2. 

— 2 X + Sy — 52=0. 

— 4x — 5>'+ i2z=o. 

Solving by ordinary algebraical methods so as to get the value 
of X we find 

X =-^ ampere. 




Fig. la. 



NETWORKS 



263 



As the generator leads are outer leads the current through 

them is of the value of the mesh cycle, or -~ ampere. This is 

also the value of the current which goes through the network 
from A to B. As the battery and its leads are taken as of no 
resistance the resistance of the network between A and B is the 
total resistance of the circuit. Call this R. By Ohm's law we 



have 



-f 



In the problem to be solved E = 2 and / 



tuting these values in this equation we have 

85 71 71 

The resistance of the network 
from ^ to J5 is 2.394 ohms. 

Example. Calculate the resist- 
ance of the same network between 
D and C, Fig. 13. 

Solution. The e.m.f. may be 
taken at i volt. Writing out the 
cycle equations by inspection we have 

3^ - 37 - oz = i. 

- Sx + Sy - 52 = 0. 

— o X — ^ y -{- 122=0. 

Solving this gives 

71 



= li 
85 



Substi- 



2.394 ohms. 




X = - — ampere. 
105 



By Ohm's law i? = — 



As E = 1, R = i 



_7i_^ 105 
105 71 
= 1.479 ohnfis. 

The value of the e.m.f. of the assumed generator may be 
taken at any number of volts desired. Thus in the first example 



264 ELEMENTARY ELECTRICAL CALCULATIONS 

it is taken at 2 and in the second at i. The value i evidently 
simplifies the operation, as the reciprocal of the value of the cycle 
of the generator mesh becomes the value of the resistance of 
the network. 

Example. Calculate the resistance of the same network 
between the points A and C. 

Taking the e.m.f. as i volt write the equations as before. 

X — y — 02 = 1. 

— X + S y — 52=0. 

— ox — 5 y + 12 z = o. 

Solving for the value of x we find x = — amperes and R = 

71 KQ ^^ 

I -4- - — = ^^ = .8300 ohm. 
59 71 ^ ^ 

Fleming's Method of Calculating Networks. In each of 
the examples given to illustrate the use of Maxwell's cycles 
we have indicated by x the cycle of the mesh in whose circuit 
the generator lies. The second member of all the equations 
except one is zero, the second member of one equation; that 
of the mesh containing the generator is the e.m.f. of the gen- 
erator. The rule for solving simultaneous equations of the first 
degree, or linear equations as they are often called, by the use 
of determinants is the following. 

As a common denominator take the determinant of all the 
coefficients of the first members, in which members all the 
unknown quantities with their coefficients must be contained and 
in which no other quantity must be present. The unknown 
quantities must be written in the same order in all equations. 
For each unknown quantity a numerator is obtained by sub- 
stituting for its coefficients in the other determinant the known 
quantities of the second member in order as written. 

Any of the groups of equations we have just given is arranged 
in order, so that determinants can be written out from them. 



NETWORKS 



265 



Take the example on page 260. Writing out the coefficients of 
the first terms of the first members we have 
6 - 2 - 



= 496 — 326 = 170. 



4 
2 8-5 
4-5 12 . 

This is the determinant common denominator and is eval- 
uated. As we wish to get the value of x or the generator cycle 
we substitute for the first or x column the terms 2, o, and 
o of the second members of the three equations. This gives us 
the determinant for the numerator. 



= (96 - 25) X 2 = 71 X 2 = 142. 



2—2—4 
o 8-5 
0-5 12 

Dividing the numerator of x, 142, by the denominator 170 
we have 

142 -^ 170 = -T— ampere = x, 
^5 

Call the determinant first written which was the determinant 
of the denominator, ^n. Then the second determinant, as it 
has two zeros in its first column, is 2. A(fj — i) and 

Aw 
But 22 = £ -V- 7 and £ = 2. Therefore 

2'A{n — i) An 



i? = 2 -^ 



An A{n — i) 

Whatever value may be assigned to the e.m.f. of the assumed 
generator it disappears in the equation which expresses the 
resistance, and we have a simple determinant expression for the 
resistance of any network. To obtain it proceed as follows : 

Assume an extra mesh of z with a source of e.m.f. connected 
as just described to the network, which mesh and generator have 
no resistance. Call this the generator mesh. 



266 ELEMENTARY ELECTRICAL CALCULATIONS 

Write out the equations of all the cycles of the network in order, 
putting all the x^s in one column, all the ;y's in another, and so on. 
When a cycle letter does not enter into the equation, insert it with 
o for its coeflScient. This has been done in the examples just 
given. The symbols of the generator mesh with their coefficients 
must be the first or left-hand members of the set of equations. 
This also has been carried out in the same examples. Write out 
the determinant of the coefficients and solve it and divide by the 
value of the first minor of its leading element. The quotient 
is the value of the cycle of the first terms; in all these examples 
this is the x cycle. 

In each of such equations there will be one term which 
has for coefficient the sum of the resistances bounding one of 
the meshes. It will be a different mesh for each equation, and 
the unknown symbol in this term will indicate the mesh for 
whose cycle the equation is written. This gives a very simple 
way of writing out the determinants directly. It is thus given by 
Fleming : 

Write as dexter diagonal of the determinant the sum of the 
resistances of each of the meshes of the network one by one, 
placing the resistance of the generator mesh in the upper left- 
hand corner. Complete the determinant by filling up each row 
with the resistances of the respective side or sides of meshes 
which separate this mesh from the one already written in its 
row. Give a minus sign to each of the latter. Where there is 
no intervening side or conductor place a zero in its place in the 

An 

determinant. Apply the rule -r—. to this determinant, 

^^ -^ A (« — t) 

and the solution will give the resistance. 

Example. Apply Fleming's rule to the network of Fig. 14. 

Each lead is supposed to have a resistance of i ; the bottom of the 

mesh >> consists of 2 leads in series, giving the bottom a resistance 

of 2. 



NETWORKS 



267 



Solution. By the rule we must start with the assumed mesh x 
and write out the dexter diagonal of the determinant. The 
sum of the resistances of the bounding conductors of the mesh x 
is 2. That of )> is 5, and z, y, u, and w have bounding con- 
ductors of resistance 3 each. It is well to write the column 
symbols in a row at the top before filling up the determinant. 
Starting with the diagonal and filling up as described we 
obtain 





X 


y 


z 


u 


w 


(I) 


2 


— I 





— I 





(2) 


— I 


5 


— I 


— I 





(3) 





— I 


3 





— I 


(4) 


— I 


— I 





3 


— I 


(5) 








— I 


— I 


3 



y 


\ 

/ 

/ 


7 


A 

z 


1 




z 

/ 




\ 


\ 


E 


\ 




2 



Fig. 14. Fig. 15- 

The minor of the leading element x of this determinant is 



= 87. 



5 - I - I o 

-I 3 o - I 

-I o 3-1 

o - I - I 3 

98 ^ 87 = 1. 126+ ohms. 



Example. Calculate by determinants the resistance of the net- 
work shown in Fig. 15 between the points A and B. 



268 ELEMENTARY ELECTRICAL CALCULATIONS 



Solution. The resistances of the meshes x^ y, z, and u are 
1,2,3, and 4 ohms respectively. These give the diagonal row or 
dexter diagonal for the determinant, which is 



o — — 



= io}. 



The first minor of the leading element of this determinant is 



22^. 



The resistance of the network between the points A and B is 
loj -H 22^ = 0.46 ohm. 

Example. Calculate the resistance of the same network 
between the points C and D. 

Solution. The imaginary mesh x is now transferred and its 
resistance is that of the lead CD, or 2 ohms, and the determinant 
of the network is 



X 

2 
o 

o 



- 2 - t 



z u 
0—2 

-i -i 

3 o 

O 4 



= 2li. 



NETWORKS 



269 



The first minor of the leading element of this determinant is 
2 -i -i 

- i 3 o = 22i. 

- i o 4 

The resistance of the network between the points C and D is 
2ii -^ 22J = 0.9663. 



PROBLEMS. 

Calculate the resistance of the network shown in Fig. 16 be- 
tween the points A and B. Each side of a mesh is of resistance 
I ohm. Ans. ij ohms. 

A 




Fig. 17. 



Fig. 18. 



Calculate the resistance of the network of Fig. 17 between the 
points A and B. Each side of a mesh is of resistance i ohm. 

Ans. f ohm. 

Calculate the resistance of the network of Fig. 18 between the 
points A and B. The resistance in ohms is marked upon the 
diagram. Ans. 1.833 or i| ohms. 



CHAPTER XIX. 
DEMONSTRATIONS BY CALCULUS. 

Force Exerted by Infinite Plane on a Point at Finite Distance. — 
Absolute Potential. — Average Value of Sine Functions. — Effective 
Value of Sine Functions. — Rate of Change. — Resistances of a 
Battery for Maximum Current. 

Force Exerted by Infinite Plane on a Point at Finite 
Distance. Let AB represent the section of a plane and m 
represent a mass of attraction m. Let the attraction of the plane 
be represented by <r for a unit of its area. Then a unit area of 
the plane at unit distance will attract the mass with an attrac- 
tion ma: At the distance r the force of attraction between the 

two will be / = -^ (i). The component of force attracting 

the mass to the plane is at any point the component perpen- 
dicular to the plane. Calling the angle between p and r, a, and 
calling the force perpendicular to the plane,/', we have,/' = 

ma- /- \ ■, • P xf ^^^P ( \ 

— ;- cos a \2)\ and smce cos a = — > / = — r- 13;. 
yi r r 

The force / and the force /' will be constant over the annulus 

whose area is 2 tt hdh. Reference to the diagram will explain this, 

w?|s^ h being the radius of 

a circle drawn on the 

plane and h + dh being 

the radius of a second 

circle concentric with 

the first, each having the 

end of p for a center. 

The annulus is included 

'^ ^ /i B between these circles. 

From the diagram we see that r^ =h^ + p^; p is a constant; 

therefore by differentiating we find 2 hdh = 2 rdr, and hdh = rdr. 

270 




DEMONSTRATIONS BY CALCULUS 2/1 

Substituting rdr for hdh in the expression 2 nhdh, we have as 
the area of the annulus 2 nrdr. The annulus is an infinitely 
small part of the area of the plane, and is therefore the 
dififerential of the area of the plane considered as a circle, 
or ds = 2 nrdr (4). The force exercised by this annulus 
at right angles to the plane is equal to the product of the 
area of the annulus by the force per unit area of the plane. 

dr 
Multiplying (3) by (4) gives f'ds = 2 nmcrp — (5). Integrat- 
ing this between the limits r = <X) and r = p gives : total force 

f'ds = 2 Tznia-p I ^ = 2 7zm<Tp 

Absolute Potential. Let e represent a charge of electricity 
concentrated at a point O. This is then a locus of attracting 
and repelling force. Let an indefinitely small body move from 
a point r cm. distant from O to a point / cm. distant therefrom. 
Whatever path the body takes in doing this the net result will 
be a movement / — r, taking / as the greater distance. 

The force acting on the body at the distance r is — • If 

moved a distance dr^ the energy will be — dr. If moved 

a distance r — /, the energy will be found by integrating this 
expression between the limits / and r. This gives 



Jr r^ \r r 



This is the energy involved in the movement of unit mass 
from r to /. The absolute potential at the point r is found by 
substituting for / the value infinity, which is equivalent to inte- 
grating between the limits r = 00 and r = r. This gives as 

the value of the absolute potential of the point r,e I — — — ) ^ " 



2/2 ELEMENTARY ELECTRICAL CALCULATIONS 

Average Value of Sine Functions. In a sine curve let 
X = Ej^^^ojt, in which t is the time elapsed since the origin of the 
curve and in which oj is the rate of angular motion and ^^ax i^ 
the radius of the generating circle. As t and oj are both referred 
to time, their product, which is a product of time and velocity, 
gives a distance which is the abscissa of the curve. In the sine 
curve it is possible to refer the motion of a point along a straight 
line to an angular velocity, because the straight line is the develop- 
ment of a circle. This line is the axis of abscissas. Let E be 
the ordinate of the curve at any point. Then the area of a 
differential element of the area inclosed by half of the sine curve 
and axis of abscissas will be Edx = £ X E^^^^wdt. The rate 
of angular motion oj, it will be seen, is a constant, and ^ is a 
variable, so that d(ot = cudt, and the differential equation is as 
above. 

By the law of the sine curve E = E^^^ sin ojt, and substitut- 
ing for E its value we have as the expression for the differential 
of the area A of the half-sine curve 

dA = £^jnax si^ (otX (o dt, (i) 

and integrating between the limits cot = n and cot = o we 
have 

A = E\,, (- cos <ot)l =- E-'^, (- I - i)= 2 E^^^, 

because cos n and cos o are each numerically equal to i. 

As it is one-half of the area that is given by this expression, the 
area of the portion above or the portion below the Hne of abscissa, 
if it is divided by the length of the line forming the base of the 
curve, which line is the portion of the axis of abscissas in length 
=nE^^^, or half the circumference of the generating circle, the 
quotient will be the average ordinate or the average value of the 
e.m.f . of an alternating system of the sine-curve type. This is 

£. . area 2 E max 2 -fcin ax ^ /:-,AA Z? 
(average) = - — = —z • = 0.6366 i^max. 

^ ^ ^ base 7r£max tt v; <« 



DEMONSTRATIONS BY CALCULUS 273 

The average value of a sine-type alternating current or e.m.f. 
is given by the above expression. 

Effective Values of Sine Functions. If a curve is laid out 
with the base line of the sine curve as its base and with the 
squares of the ordinates of the sine curve as its ordinates, and if 
the area of such curve is divided by the base line, the quotient 
will be the average value of the ordinates, which is the average 
value of the squares of the ordinates of the sine curve. The 
square root of this quantity is the square root of the average 
value of the squares of the sines or ordinates of the sine curve or 
values of the alternating current. It is the effective value of the 
current. 

The differential of the area of the new curve is given by the 
product of the square of an ordinate P by the differential of the 
base line. The base line is equal to /^^x ^^^ i" radian measure- 
ment, so for the differential we have 

dA=PX /„,,x dcot, (i) 

and substituting for i its value in terms of /max) which is /^ax 
sin ojt, (i) becomes 

dA = /^max sin^ cotdcot, (2) 

From trigonometry it is known that 

. „ ^ I — cos 2 (Ot 
sin'' cut = > 

2 

and substituting this value in (2) and multiplying both numer- 
ator and denominator by 2 we find that we obtain a readily 
integrable expression : 

J . T, 2 dcot — 2 cos 2 cotdcDt , . 

^^ = ^'max ' (3) 

4 
whence by integrating between the limits (ot = n and cot = o 
we have 

. ^o [2 (ut — 2 sin ojff , ^ 



2/4 ELEMENTARY ELECTRICAL CALCULATIONS 

The value of n is i8o°, whose sine is o; the sine of o is also o, 
giving 

A — Ji ^lJL — ^^ max . ,. 

^ - ^ max - 2 ^5) 

The base of the area A is nl^^^. Dividing (5) by this value 
gives the average value of the ordinates of the area, 

-^ ^ m ax ,ri\ 

'^^ max ^ 

and the square root of the second member of this equation is the 
square root of the average square of the ordinates of the sine 
curve, which is the effective or virtual current, or 

/max 
/ef = —7= = 0.707 /max. (7) 

Rate of Change. The amount of change in the value of a 
sine current at any place is given by the difference between the 
values of two consecutive ordinates of the curve at that place 
multiplied by any coefi&cient required. The relative amount of 
change is then expressed by {y ^ dy) — y = dy. The time dur- 
ing which this change takes place is the difference of the corre- 
sponding abscissas, or {x + dx) — x = dx. As the differentials 
are those of current and of time they may be written di and dt. 
The rate of change is the quotient of change in value of current 
divided by the time required for such change to take place, 
or di/dt. 

Let (o be the angular velocity of the current vector, which is 
the radians per second covered by the current alternations and in 
terms of frequency is equal to 2;r/, because a cycle is equal to 
2 7r radians. The value of the current at any instant is given by 
the expression /max sin 6. But 6 is equal to cot, 6 being the 
degrees corresponding to t radians. Introducing this value of the 



DEMONSTRATIONS BY CALCULUS 275 

angle of position in the expression for the current value we 
have the equation 

i = /max sin cot, 

in which i is the current value at any time cot. 

Differentiating this expression with respect to t we have 

di = /max cos Ojtdcot = culmsix COS (Otdt, 

and dividing by dt, 

di 

— = ojlmax COS cot = rate of change at t, 

and substituting for co its value, 2 tt/, and for cut its value in degrees, 
6, we have 

Rate of change at ^ = 2nfI^^^cosd. 

Resistance of a Battery for Maximum Current. Let n 

be the total number of cells in a given battery, and let r be 

the resistance of a single cell, and x be the number of cells 

n 
in series. Then - will be the number of cells in parallel, and 

X 

ft fX^ 

by Ohm's law the resistance of the battery will he rx -^- = — ■ . 

X n 

Let R be the resistance of the outer circuit, and e be the e.m.f. 

of a single cell. Then the e.m.f. of the battery will be ex, 

R -\ ) = ^ — r • The 

n / nR -\- roc 

value of R which will make this quantity a maximum is to 

be determined. The operation is simplified by using the 

reciprocal and finding the value of R which will make the 

iiR + rx^ 

reciprocal a minimum. It is obvious that if an 

nex 

expression is a maximum, its reciprocal is a minimum. We 

then reduce the expression and omit the factor - before differ- 

e 



2^6 ELEMENTARY ELECTRICAL CALCULATIONS 

entiating and then proceed by the regular process of maxima 
and minima, 

nR -{- rx^ _ ^ f^ ^A 
nex e\xnl^ 

'R rx\ nrdx Rdx 



/K rx\ nrdx Rdx 
\x n I n^ 0^ ^ 



and taking the differential coefficient equal to zero, we find 

the value of R which makes the reciprocal of the current a 

^ r R ^ ry? 

maximum or a mmimum ; thus 5=0 and R = 

n or n 

To determine whether this value of R gives a maximum or 

minimum value of current we must find the sign of the second 

differential coefficient. Proceeding by the regular methods 

find d \ r) =2 Rxdx, giving the second differential 

\n xrl 

coefficient, 2 Rx^ which is positive. Therefore the value 

ry? 
R = — gives a minimum value to the reciprocal of the cur- 
n 

rent or a maximum value of current. But we have seen 

that — is the internal resistance of the battery. Therefore 
n 

the arrangement of cells which will make the external and 

internal resistance equal will give the maximum current. 



we 



APPENDIX A. 



GEOMETRICAL SOLUTION OF PARALLEL CIRCUITS. 

The resistance of two conductors in parallel can be deter- 
mined by a geometrical construction. 

On the extremities of any line as a base erect two perpendicular 
lines, a and 6, proportional to 
the resistance of the two conduc- 
tors. Draw diagonals as shown 
from the upper end of each line 
to the base of the other. 

From the point of intersection 
of the diagonals drop a perpen- 
dicular c to the base line. The length of c is proportional to 
the combined resistance of the two conductors. 

For this to be true, the value of c must be given by the follow- 
ing equation: aXh 




c = 



(see page 65). 



a + b 
This is proved as follows: 

Let c divide the base line into the two parts m and n. 

a _ m -^ n 
c n 

b m -\- n 



c 
a 
b 

a + b 



m 
m 
n 

m + n 



Dividing (i) by (2) 

b n 

by composition substituting from (i) 

g + 6 _a 

b T 

ab 



which gives 






a + 6 
277 



Then 

(I) 
(2) 
(3) 
(4) 

(s) 

(6) 



Q.E.D. 



2/8 ELEMENTARY ELECTRICAL CALCULATIONS 

If the resistance of three or more conductors in parallel is 
required the same method can be applied. 

Erect on the horizontal line a perpendicular line for each of 
the resistances, the lengths of the lines being proportional to 
the resistances which they severally represent. Combine any 
two of them by the method just described. Then, following 
exactly the same method, combine the resistance represented 
by the new perpendicular with that of one of the remaining 
perpendiculars. 

This gives the combined resistance of three conductors. 
The process is followed out until the resistances have all been 
used. The final result will be a perpendicular line much 
shorter than any of the constituents and whose length will be 
proportional to the combined resistance of all the resistances 
represented by the perpendiculars originally drawn. 

As is the case with many other graphic methods, the accuracy 
of this process is not very great, and it naturally becomes less 
accurate as more resistances have to be combined. This is 
because the line proportional to and representing the com- 
bined resistances^ soon becomes too short for accurate 
measurement. 



1 



APPENDIX B. 

ALGEBRAIC SOLUTION OF CIRCUITS. 

Elementary Cases of Net Works. Some typical circuits are 
treated algebraically in the following pages. By the applica- 
tion of Kirchhoff's and Ohm's laws and the use of algebra, the 
formulas for their solutions are deduced. They can also be 
solved by MaxwelFs cycles, as explained in Chapter XVIII. 

Example. Calculate the distribution of ^ 

currents in the circuit shown in the dia- /" " 

gram. :=: (3 

Solution. The circuit is treated as of V^ 




three branches, i, 2 and 3. Branch i in- ° 

eludes the generator. Subscript letters refer to the respective 
branches, r indicating resistance and i indicating current. The 
combined resistance of 2 and 3 is — ^— ^ (see page 65) (i). 

^2 + n 

The total resistance of the circuit is 

^2+^3 ^2 + ^3 

Let E be the e.m.f. of the generator. By Ohm's law {I=E/R) 

i = E -r- ^^^2 + ^2^3 + ^ih ^ ^2 + ^3 y^ £ /^\ 

^2 + ^3 ^/2 + ^2^3 + ^1^3 

The e.m.f. expended in each of the two branches, 2 and 3, is 
identical in amount by Kirchhoff's law\ It is equal to the com- 
bined resistance of the two branches divided by the total resist- 
ance, the quotient being multiplied by the total e.m.f. of the 
system. This gives 

e^ or ^3 = ^^2-il3 XE= -^^^^ ^ ^i^ + ^2^3 + r,r^ y^ ^ 
R ^2 + h ^2 + h 

^1^2 + ^2^3 + ^1^3 
279 



28o ELEMENTARY ELECTRICAL CALCULATIONS 

By Ohm's law i^ = ^ (5), and 4 = ^ (6). 

Substituting in (5) and (6) the values of e^ and e^ from (4) 
gives 

h = — ^^— X E (7) and 4 = —^ (8) 

These expressions for the resistances of the branches 2 and 3 

have the same denominators, therefore they have the ratio of 

j^ their numerators, which ratio is the inverse 

^ 7^ "\ ratio of the resistances of the branches, or 

^^ Vly Example. Assume three branches, 2, 3 

D 

and 4, starting from A, and coming to- 
gether at B. Calculate the currents. 

Solution. Proceeding as before, we have R =fi+ (r^; r^; r^) = 

^2^3 + ^2h + ^3^4 ^2^3 + ^2^4 + ^3^4 

This is the total resistance of the circuit. The distribution 
of current has now to be determined. Proceeding as before, 

e„e„ore„ = Ex'-^-^^^ = ^X -^^^ ' (lo) 

R R r^r^ + r^u + hU 

and substituting for R in this equation its value from (9), and 
reducing by carrying out the indicated operations, we obtain 

e„ e„ or e,, = E X ■ ; ^-^ (n) 

As L = ^ (12), % = ^ (13), and i^ = ^ (14), (11) divided by 
^2 ^3 ^ 

^2, ^3 or r^ will give the respective currents in branches 2, 3 
and 4. Thus for branch 2, divide (n) by r^, giving: 

'-^=i,= EX——- '-^ (15) 



APPENDIX B 



281 



For branches 3 and 4 the values of the currents would differ 
only in the numerators; for branch 3, the numerator would be, 
r^r^y and for branch 4, the numerator would be, r^r^. 

Example. Calculate the distribu- 
tion of current in the circuit shown 
in the diagram. 

Solution. The branches i and 2 
include generators. The symbols 
correspond with those of the preced- 
ing example. 

The e.m.f. of the generator of branch 2, expended in branches 
I and 3, taking account only of the resistance, and omitting 
consideration of the counter e.m.f., is 




gj or ^3 (of generator 2)= £3 X ^1 ; ^'3 -^ (^2 + ^1 j ^3) = 



£2X 



^1 + ^3 ' ^1 + H 



E,X 



JVs 



•(I) 



^1^2 + hh + ^2^3 

Going through the same operation for the e.m.f. in branches 
2 and 3, due to generator i, we obtain 

^2 or ^3 (of generator i) = Ej X ^3; r^ ^ {r^ -\- r^', r^) giving, as 
above, _ _ _ ^y 



E^X 



r,r^ + r, r^ 



(2) 



If the two generators are arranged as indicated in the diagram, 
they will work together as far as the branch 3 is concerned, and 
the e.m.f. of 3 will be the sum of the second members of (i) and 



(2), or 



E,r^r^ + E,r,r^ 

h^2 + ^1^3 + ^^3 



(3) 



Applying Ohm's law, % = -^ , divide the above equation by 



the resistance of 3, which is r^. This gives 

i = fa = ^^^■> + -^^^1 . 
^3 h^2 + hh + ^2^3 



(4) 



2^2 ELEMENTARY ELECTRICAL CALCULATIONS 



If the generators are in opposition as regards branch 3, the 
current in 3 will be 



n^2 + ^1^3 + ^2h 

The e.m.f. of branch 2, due to its own generator, is 



62 (produced by generator 2) 



r'> (r, + r^) 



The e.m.f. of branch 2, due to the generator i, is given by 
equation (2). Subtracting the last member of (2) from the last 
member of (5) gives the actual e.m.f. of the branch, 5, because 
one generator as shown is working in opposition to the other. 
Then, applying Ohm's law, 






(6) 



Exactly similar processes give the value of the current in i : 

l^ ^e,_ E, (r.+ r,)- E.r^ ^^^ 

Had one of the batteries or generators been connected with 
opposite polarity to that shown, equations (6) and (7), with a 

plus sign in place of the minus 
sign between the two terms of 
the numerator, would give the 
value of the currents. 

Example. Calculate the cur- 
rents in the'^Wheatstone bridge" 
fg circuit, shown in the diagram. 
Solution. Let the arrows de- 
note the direction of the cur- 
rents. By Kirchhoff' s first law 




to = ii 



(I); i, 



+ h< (2); 'e 



APPENDIX B 283 

By Ohm's law the e.m.f. of branches 3 and 6 is obtained by 
multiplying both members of (i) and (3) by r^ and Tq respec- 
tively. 

rzh=hh-hh (4); V6=Vi+^6V (5) 

The entire e.m.f. of the system is expended on the circuit i, 3 
and 6. Adding the second members of (4) and (5) to r{l^, the 
latter, the e.m.f. of i, gives the total e.m.f. of the system, because 
such sum is the sum of the e.m.f. 's of the three branches, i, 3 
and 6. Thus: 

£=^3^'i-%+ V1 + V2 + hh^hih + ^6+ ^1) + ui^-r^H- (6) 
The e.m.f. expended in the branch 2 is equal to the sum of 
the e.m.f. 's of i and of 5, giving 

hh = hh + hh or ^ih - ^2h + hh =^ o- (?) 

The e.m.f. of 3 is given by adding that of 4 to that of 5, or 

hh = hh + hh' (8) 

Substituting for r^i^ in this equation, its value from (4), gives: 

hh - hh = Uh + vV (9) 

The current in 4 is equal to the sum of the currents in 5 and 
2, giving as in the other parallel processes 

h = h + h^ (10) 

and r^i^ ^ rj^^ + rj^. (11) 

Substituting this value of rj^ in (9) gives 

rJi - f'sU = ''4^2 + fih + hh, (12) 

and transposing and factoring, 

hh - Uh - h (^3 + ^4 + ^5) = o. (13) 

In (6), (7) and (13) three currents appear, i^, i^ and %• We 
next proceed to ehminate i^ and ^2 from them, treating them as 
simultaneous equations. 
Multiply (7) by r^. This gives 

^>/3 - ^>2^3 + hhh = o- (14) 



284 ELEMENTARY ELECTRICAL CALCULATIONS 

Multiply (13) by r^. This gives 

Ws - Wi - h (^/3 + r,r, + r,r^) = o. (15) 

Subtracting (14) from (15) gives 

h (^2^3 - ^1^4) - h ihh + hU + r^h + hh) = o. (16) 

Dividing by {r^r^ — r-j^ and transposing, gives 

Substitute this value of ^2 in (7) and transpose. 

^2^3 - r,r. 
Divide both members by r^, and bring — r^i^ into the numer- 
ator by multiplying it by r^r^ — r^r^. This gives 

^2^3 + ^2^4 + ^2^5 + -^-^-^ - -^-^ + ^4^5 

ii = h . (19) 

^2^3 - hr. 

The fourth and fifth terms of the numerator cancel each 

other. Substituting the values of ^3 and i^ from (17) and (19) in 

(6), we obtain 

^2^3 - ^1^4 
+ ^5 -^-^ — -^ -^ ^6 - HH' (20) 

Bring r^\ into the numerator by multiplying it by y.^y^ — r^r^, 
perform the indicated multipHcations, and arrange the terms of 
the numerator in numerical order. The numerator then becomes 

^1% + r,r,r, + r,r,r^ + r,r,r^ + ^^^3^3 + ^2^3^4 + ^2^3^-5 + 

''3^4^5 + ^2^3^6 + ^2^4^6 + ^-2^^6 + ^1^6 + hhH + hU^% + ^1^5^6 + 

In this expression the fifth and seventeenth terms cancel 
each other, and by factoring the expression becomes 

^5^6 (^1 + ^2 + ^3 + ^4) + H (n + h) (^2 + U) + ^6 (Ti + h) 
(h + h) + hh (^2 + U) + ^2^4 (^1 + ^3). 

which expression we will call D. 



APPENDIX B 



285 



Then write out (20) as modified, using the symbol D to 
express the numerator, and we have 

D 



E = L 



hh - hn 



(21) 



dividing by 



D 



r^r^ — r,r^ 



and transposing, 



(22) 



(23) 



'2' 3 '1'4 

' D 

This is the current in branch 5. 
Transposing (3) gives 

^2 ^^ ^6 H' 

Substituting this value of ^'3 in (7) and factoring, gives 

hh - ^2^ + ^2h + hh = h in + ^2) + hh - ^2h = o. (24) 
Doing the same in (13), gives 

hh - (h - h) U - H (^3 + U+ h) = h ih + ^4) 

- H (h + U+ h) - hh = o- (25) 

Eliminate i^ from (24) and (25), by first transposing and 
dividing (24) by r^ + ^-2, and (25) by r^ + r^ respectively, which 
gives 

i ^ rA^LlA (,6), and i, = ^^^« + ^'^ ^^b + ^ + ^0 , ( ) 
h + r, ^3 + ^4 

and then complete the elimination by equating the second mem- 
bers: 

Clear of fractions, perform the indicated multiplications, and 
factor. The term ^2^4^ cancels out in the operations, and 
transposing we obtain 

h (^2^3-^iO = hh (^1+^2 + ^3 + ^4) + h (Xi + ^2) {h + O; (29) 
whence we obtain 

^- _ ^'■'^ r^s (^1 + r, + r, + r,^ + (fi + ^) jr^ + r,)1 ^ . . 



286 ELEMENTARY ELECTRICAL CALCULATIONS 

Substituting for 4 its value from (22), the binomial r^^ — 
r^^ disappears, and we obtain 

l^ _ £ [r, {r, + r, + ^3 + r,) + {r, + r,) (r, + r,)\ . ^^^^ 

This is the current in branch 6. 

Substitute in (26) the values of ^ from (22), and of \ from 
(31). 

^(^'+:^\ . _ (3^) 

Performing the indicated multiplications, factoring, and 
reducing, with the disappearance of the terms r^Tzr^ and r^ + ^'2 
we obtain 

This is the current in the branch i . 

From equation (3) we find, by transposing, 

h = H - hy (34) 

and substituting for Iq and i^ their values from (31) and {2,^), 
performing all indicated operations, simpHfied by the fact that 
both fractions have the same denominator, we obtain 

.^ _ E {r, r^ + r,r^ + rf^ + r , r , )^ ^^^^ 

This is the current in branch 2. 

Equation (i) reads 4=%— %• Substituting in it for ij and i^ 
their values from (22) and {t,:^,), and reducing as hitherto, we find 



= £ (^?^ + ^2^5 + ^^5 + ^1^4) 



This is the current in branch 3. 



(36) 



From equation (2), i^ = i^ + H', substituting the values of 
^2 and 4 from (22) and (35), and reducing, we find 

This is the current in branch 4. 



APPENDIX C. 



WHEATSTONE BRIDGE LAW. 




(I) 
(2) 

(3) 



The Wheatstone bridge law may be thus deduced by the use 
of determinants. The usual diagram represents the Wheat- 
stone bridge connections, and in it capi- 
tal letters indicate the currents flowing 
through the arms of the bridge and 
through the galvanometer connection, 
and the small letters indicate resist- 
ances of the same parts. 

From Kirchhoff's first law we have the 
three relations 

R = P-G, 

S=Q-hG. 

These equations refer to the currents. 

The e.m.f. at different points of the bridge is given by Kirch- 
hoff's second law, giving the three equations 

Gg+Pp-Qq = o, (4) 

-Gg +Rr -Ss = o, is) 

Bb + Qq -}- Ss ^ E. (6) 

in which last equation E represents the e.m.f. of the battery. 
Substituting the values of B, R, and S in equations (4), (5), 
and (6) from equations (i), (2), and (3) we have 

Gg + Pp-Qq= o, (7) 

-Gg+ {P-G)r- iQ + G)s = o, (8) 

(P + Q)b+Qq+ iQ + G)s = E, (9) 

287 



288 ELEMENTARY ELECTRICAL CALCULATIONS 



These we may rewrite as 

Gg + Pp- Qq = o, (lo) 

- G{g + r + s) + Pr - Qs = o, (ii) 

Gs + Pb+ Q{b + q+s) = E. (12) 

Solving this by the regular determinant method we have for 
the general divisor determinant 

P 
r 
b 



- q 

— s 

b + q + s, 

whose value we may indicate by D. The numerator determi 
nant for the value of G : 



- (g+r+s) 
s 



o P - q 

r — s 

1 b (b-h q+ s). 

The evaluation of this determinant is qr — ps, giving us for 
the value of the galvanometer current, G, 



^ E{qr - ps] 



D 



(13) 



From equation (13) it follows for G to be equal to zero that qr—ps 
must be equal to zero, or qr = ps, which put in the form of a 
proportion gives the proportional form of the law of the Wheat- 
stone bridge, 

q : p \ \ s '. r. 



TABLES OF EQUIVALENTS 





Length. 


I in. 


= 


25.40010 mm. 


I ft. 


= 


0.30480 Meter 


I yd. 


= 


0.91440 Meter 


I mile 


= 


1.60935 Km. 


I Nautical Mile 


= 


1853.25 Meters 


I fathom 


= 


1 . 829 Meters 


I Meter 


__ 


39.37043 In. 


I Meter 


= 


3.28083 Ft. 


I Meter 


= 


1. 09361 Yds. 


I Km. 


= 


0.62137 Mile 




Area. 


I Sq. In. 


= 


6.452 Sq. cm. 


I Sq. Ft. 


= 


9.290 Sq. dm. 


I Sq. Yd. 


= 


0.836 Sq. M. 


I Sq. Mile 


^ 


259.008 Hectares 


I Sq. cm. 


^ 


0.1550 Sq. In. 


I Sq. M. 


= 


10.764 Sq. Ft. 


I Sq. M. 


= 


1. 196 Sq. Yd. 




Weight. 


I grain 


= 


64.7989 mg. 


I oz. Av. 


= 


28.3495 Gm. 


I oz. Troy 


= 


31.10348 Gm. 


I lb. Av. 


= 


0.45359 Kilo. 


I lb. Troy 


= 


0.37324 Kilo. 


1 lb. Av. 


— 


453-5924 Om. 



I mg. = 0.01543 gram 

I Gm. = 15.43236 grains 

I Kilo. = 33.814 flu. oz. 

I Kilo. = 2.20462 lb. Av. 

I Kilo. = 2.67924 lb. Troy 

I Kilo. = 35.274 oz. Av. 

I Kilo. = 32.1507 oz. Troy 

I Millier or Tonne = 2204.62 lb. Av. 

I Quintal = 220.462 lb. Av. 
289 



290 ELEMENTARY ELECTRICAL CALCULATIONS 



Volume. 

I minim (water) = 0.06161 c.c. 

I flu. dr. = 3.70 c.c. 

I flu. oz. = 29.5737 c.c. 
I Apoth. oz. (water) = 31.10348 c.c. 

I quart = 0.94636 Liter 

lU. S. gal. = 3.78543 Liters 

I bushel = 0.35239 Hectol 



I c.c. 
I c.c. 


= 16.23 minims (water) 
= 0.2702 flu. dr. 


I Centiliter 


= 0.338 flu. oz. 


I Liter 
I Liter 
I Decaliter 
I Hectoliter 


= 1.0567 qt. 
= 0.26417 gal. 
= 2.6417 gal. 
= 2.8377 bushels 


I cu. in. 


= 16.387 c.c. 


I cu. ft. 


= 0.02832 c. M. 


i cu. yd. 


= 0.765 c. M. 



I c.c. 
I c. dm. 
I c. M. 
I c. M. 



I Poundal 
I Pound 
I Grain 
I Gram 



= 0.06102 cu. in. 
= 61.023 cu. in. 
= 35-314 cu. ft. 
== I -308 cu. yd. 

Force. 

= 13,825 dynes 

= 4.45 X 10* dynes 

= 63.6 dynes 

= 981 dynes 



I foot-pound 
I foot-poundal 
I foot-ton 
I joule 



Energy. 
= 13,823 gram-centimeters 
= 3.096 X 10' gram-centimeters 



1.3560 X 10' ergs 
4.214 X 10* ergs 
3.0374 X 10^" ergs 



= , lo" ergs 

Power, Energy Rate, or Activity. 



I horse-power = 746 watts 

I horse-power = 7.604 X 10' gm. cm. per second 7.46 X 10® ergs per 

second 
I metric horse-power = 7.5 X 10' gm. cm. per second 7.36 X 10' ergs per 

second 
I kilowatt = 10^" ergs per second 

I watt =16' ergs per second 



I 



TABLES OF EQUIVALENT^ 



^9t 



Mechanical Equivalent of Heat. 

I gram-degree C. = 4.281 X 10* gm. cm. 4.2 X 10' ergs 

I pound-degree F. = 1.942 X 10' gm. cm. 1.905 X io^° ergs 

(B.T.U.) 

I pound-degree F. =777.4 foot-pounds. 



I U. S. gal. 

I U. S. gal. (water) 

I bushel 

1 lb. Av. (water) 

1000 grains (water) 

I pint 

I flu. oz. 

I flu. oz. (water) 

I oz. Av. (water) 

I lb. Av. (water) 

I cu. in. (water) 

I liter (water) 

4 c.c. = I dram 

8 c.c. = 2 dram 

16 c.c. = 4 dram 



Miscellaneous. 

= 231 cu. in. = 4 qts. = 8 pints 

8.313 lb. Av. = 58,418.1444 grains 
2150.42 cu. in. 
0.12029 gal. 
2.200 flu. oz. 
16 flu. oz. 

8 drams = 480 minims 
456.392 grains = 1.0391 oz. Av. 

0.9623 flu. oz. 

0.12029 gal. 
252.892 grains 
I Kilo. 
I teaspoonful 
I dessert-spoonful 
I table-spoonful 



i 



INDEX 



PAGE 

Absolute and practical units ii6 

C.G.S. electro-magnetic unit of current io6 

C.G.S. electrostatic unit of quantity 112-114 

electric potential 202, 203 

temperature 137 

Acceleration 20 

of gravitation 22 

Action of batteries , 82-93 

Activity 24 

or power, electric 97 

Addition, algebraic 4 

Addition and subtraction of powers 13 

Algebra 4 

Algebraic addition 4 

solution of circuits 279-286 

subtraction 5 

symbols 4 

Alternating current 220, 221 

currents, induction of 220, 221 

electro-motive force, induction of 220 

Ampere 6 

Ampere turns 186, 187 

for a given field 191 

Angle of divergence 109 

of lag and lead 236 

of lag and rate of change 242, 243 

Arithmetic, short methods in i 

Atomic weights 144 

Batteries, discussion of principles of calculating 88-90 

Battery, electro-motive force of 82 

energy expended in a 86 

greatest current from a 83 

potential drop of a 83 

resistance of a 82 

resistance of, for maximum current, by calculus 275 

293 



294 INDEX 

PAGE 

Battery, rules for calculating a 84-86 

rules for calculating a, of given eflQciency 86 

British system of units 25 

Calculating batteries, discussion of principles of 88-90 

Calculus 6, 268-273 

Cancellation i 

Capacity 198 

and inductance, combined impedance of 234, 235 

and potential, heat analogy of 204 

inductance and resistance impedance of 236 

measure of 198 

of parallel plates 199 

ohmic equivalent of reactance of 233, 234 

potential and quantity, relation of 202, 203 

specific inductive 201, 202 

Centimeter-gram-second system 18 

Central fields 159 

force ss 

C.G.S. electric units, two systems of 105 

electro-magnetic unit of current, absolute 106 

electrostatic unit of quantity, absolute 11 2-1 14 

units of e.m.f. and resistance, how determined iio-iii 

unit turns 186, 187 

Change, rate of 228 

energy of a .• 201 

Chemical composition 142 

equivalent 144 

equivalents and electrolysis 146 

saturation 143 

Chemistry, thermo-electric 149 

Circuit, factors of an active 41 

magnetic, calculations 189-197 

the magnetic 185, 186 

Circuits, algebraic solution of 279-286 

geometric solution of parallel. 277, 278 

Circular mils 7c>-72 

Coefficients 5 

Coil, field in 188 

Compass, tangent 108 

Composition, chemical 142 

Conductance of parallel conductors 64 

Conductor, heating of, by a current. . 126 



INDEX 295 

PAGE 

Conductor, size of, for a given current 126-131 

Conductors, linear 53 

diameter of, resistance of same 53-55 

parallel 58-68 

parallel, conductance of 63 

parallel, resistance of 63-68 

resistance and weight of linear 56-58 

resistance of square and circular 55 

resistance of two, in parallel 65 

temperature of, and resistance 73 

Coulomb 115 

Couple 167 

thermo-electric 131 

voltaic, e.m.f. of 149-152 

Counter electro-motive force 49 

and inductance 207 

Current 6 

absolute C.G.S. electro-magnetic unit of 106 

alternating, induction of 220, 221 

basis for measurement and definition of a 106 

change, rate of, and the henry 206 

change, variation of rate of 207 

dimensions of, in E.M. system 120 

direction of, in networks 248, 249 

distribution of, in parallel conductors 59~63 

effects of an electric 105 

electrostatic unit of 115 

greatest, from a battery 83 

heating of a conductor by a 126 

induction of 161-164 

instantaneous, value of 223 

lag of 232, 233 

lead of 235 

resistance and e.m.f., relations of power to 98 

size of conductor for a given 126-131 

Currents, eddy, formulas for, in wire cores 215 

eddy, formulas for, in laminated cores 213 

Foucault, formulas for, in wire cores 215 

Foucault, formulas for, in laminated cores 213 

in networks 254, 255 

Cycle 223 

calculations in networks 252, 253 

equations, Maxwell's rule for 257 



296 INDEX 

PAGE 

Cycle, letter in networks 254 

Cycles in networks 247 

how indicated 251 

Dielectric constant 202 

power 202 

Dimensions, E.M. system, of capacity, 123; of current, 121; of elec- 
tric quantity, 122; of magnetic quantity, 121; of potential, 

122; of resistance 122 

Dimensions, E.S. system, of capacity, 124; of current, 124; of 

electric intensity, 125; of electric quantity, 123; of potential, 

123; of resistance, 124; of surface density of electricity .... 123 

Dimensions, of magnetic intensity, 124; of magnetic potential, 

125; of magnetic power, 125; of magnetic quantity, 124; of 

surface density of magnetism 124 

Dimensions of mechanical units 35~38 

Dimensions, theory of 35 

Divergence, angle of 109 

Division, indication of 2 

of different kinds of units 17 

Drop, potential 47> 95 

potential, of a battery S^ 

RI 47, 95 

Dyne 12 

Earth's field, action of 109 

Eddy currents 213 

formulas for, in laminated cores 213 

formulas for, in wire cores 215 

Effective value of sine functions 225, 226 

Effect, Peltier 137 

Thomson 138 

Efficiency 32 

rules for calculating a battery of given 86 

Electric current, effects of an 105 

Electric decomposition . . . -. 144 

Electric energy 9^ 

and power 94-104 

practical unit of 97 

Electric potential, absolute 202, 203 

Electric power or activity 97 

practical unit of 97 

Electric units, bases and relations of 105-125 

two systems of C.G.S 105 



INDEX 29; 

PACK 

Electro-chemical equivalents 146 

Electrolysis 144 

Electrolysis and chemical equivalents 146 

Electro-magnetic field of force, energy of the 208 

Electrometer, attracted disk 11 2-1 14 

Electro-motive force, alternating, induction of 220 

counter 49 

instantaneous, value of 223 

of a battery 82 

Electrostatic system, current unit 115 

resistance unit 115 

unit of quantity, absolute CG.S 11 2-1 14 

E.M. and E.S. systems of units 11 5-1 16 

E.M. and E.S. units, reduction factor of 119 

E.M.F. and resistance, determination of CG.S. units of iio-iii 

how calculated 149-152 

induction of 161-164 

resistance and current, relations of power to 98 

Emissivity 126 

E.M. practical units directly derived 118 

E.M. system, dimensions of current in 120 

Energy 23 

and power, electric 94-104 

available 23 

electric 96 

expended in a battery 86 

heat 29-30 

kinetic 24, 27 

of a charge 201 

of the electro-magnetic field of force 208 

potential 24 

practical unit of electric 97 

units and equivalents 30 

varieties of 24 

Entropy 23 

Equivalence of units 7, 28 

Equivalents, chemical 144 

electro-chemical 146 

of the watt 100 

Erg 24 

E.S. and E.M. systems of units 115 

Exponential notation i, 8-15 

Exponents 8 



298 INBEX 

PAGE 

Exponents, changing 12 

decimal 9 

factoring 13 

fractional. 9 

negative 10 

Factor, form 227 

power 239—240 

Factoring exponents 13 

Fall of potential 47 

Farad 116, 199 

Field, action of earth's 109 

and ampere turns ^ 191 

in coil 188 

intensity of 155 

of force, energy of the electro-magnetic 208 

polarity of 156 

quantity of 156 

strength of 187 

the unit . 155 

total 188 

Fields, central 159 

of force 155 

of force, reciprocal action in 159 

uniform and of varying strength 158 

radiant 159 

Filament, lines of force in a magnetic 167, 168 

Fleming's method of calculating networks 264-266 

Foot-poundals 26 

Force 21 

central or radiant. . t,^, 

exerted by infinite plane on a point at finite distance 270, 271 

fields of 155 

lever arm of 167 

lines of ~. 155-156 

lines of, in a filament 167 

magneto-motive 188 

of a plane 34-35 

reciprocal action in fields of 159 

Form factor 227 

Foucault currents 213 

formulas for, in laminated cores 213 

formulas for, in wire cores 215 



INDEX 299 

PAGE 

Frequency 223 

Functions 5 

Fuses, safety, calculating 130 

Galvanometer, tangent, formula of, deduced 1 09-1 10 

Geometrical solution of parallel circuits 277 

Gravitation, acceleration of 22 

Gravitation, weight and 18-19 

Heat energy 29-30 

Heating of conductors by currents 126-134 

Henry, the 206 

Hysteresis and energy 211 

formulas, Steinmetz's 211— 212 

formula, volume 211 

formula, weight 212 

Hysteresis loss 211 

Hysteresis table, Steinmetz's 212 

Impedance 234, 235 

of inductance and resistance 231 

of inductance, capacity and resistance 236 

Index number 8 

Inductance 205 

and capacity, combined impedance of 234, 235 

and resistance, impedance of 231 

capacity and resistance, impedance of 236 

reactance of 227-229 

reactance, ohmic equivalent of 229 

Induction, magnetic 180 

alternating current 220, 221 

of alternating electro-motive force 220 

of current 161-164 

of e.m.f 161-164 

of magnetism 178 

Inductive capacity, specific 201, 202 

Instantaneous values 223 

Intensity of field 155 

of magnetization 172, 174 

loule 29 

Kapp's unit 158 

Kinetic energy 24, 27 

Kirchhoff's laws 77-8i 



300 INDEX 

PAGE 

La^ 231, 232, 235, 236 

and lead, proof of the law of the angle of 237-239 

angle of, and rate of change 242, 243 

or lead, angle of 236 

of current 232, 233 

or lead due to reactances 235 

Law, chemical, of multiple proportions 142 

Law, Ohm's, see Ohm's Law. 

Laws, Kirchhoff's 77-8i 

Lead 231, 232, 235, 236 

and lag, proof of the law of the angle of 237-239 

of current 235 

or lag, angle of 236 

or lag, due to reactances 235 

Lever arm of a force 167 

Linear conductors 53 

Lines of force 155-156 

and magnetic quantity 170 

in a filament 167 

Logarithms 2 

Magnetic circuit 185, 186 

circuit calculations 189-192 

filament 167 

induction 180 

quantity 169, 1 70 

quantity, dimensions of 120 

traction 192-194 

Magnetism, induced, and field 178 

induction of 178 

Magnetization, intensity of 172-174 

Magnet, moment of a 171, 174 

Magneto-motive force 188 

Mass 18 

Maxwell's rule for network equations 257 

Meshes in networks 247 

outer and inner, in networks 250 

Microfarad . . . , 199 

Mils, circular 70-72 

Moment of a magnet 171, 174 

Moments 166 

Moment, unit of 167 

Multiple proportions, chemical law of 142 

Multiplication of different kinds of units 17 



i 



INDEX 301 

PAGB 

Negative, positive and , 4 

Networks 247-268 

current in 252, 255 

Fleming's method of calculating 264 

resistance of 262, 263 

n meaning of 3 

Notation, exponential i, 8-15 

Numerical values 7 

Ohm 6, 1 16 

Ohmic equivalent of capacity reactance 233, 234 

equivalent of inductance reactance 229 

equivalent of reactance .' 227-230 

Ohm's Law, 41-52; three forms of, 41; several appliances in one 
circuit, 42; application to portions of a circuit, 43; simple 
method of expressing, 44; proportional forms of, 45; fall of 
potential, 47; RI drop, 47; counter electro-motive force, 49; 
generators in opposition 50 

Parallel circuits, geometric solution of 277, 278 

conductors 58-68 

Peltier effect 137 

Permeability 181, 182 

tables of 183 

Permeance 182-185 

Permitivity 202 

Perviability 202 

Plane, force of a 34~35 

Polarity of field 1 56 

Pole, strength of, and magnetic quantity 169 

Positive and negative 4 

Potential 94 

absolute electric 202, 203 

and capacity, heat analogy of 204 

capacity and quantity, relation of 202, 203 

difference and quantity 203, 204 

drop 47, 95 

drop of a battery 83 

energy 24 

fall of 47 

Potential, proof of numerical value of 94 

Poundal 25 

Power 24 

and energy, electric 94-104 



302 INDEX 

PAGE 

Power and power factor , 239-240 

factor and reactances 241 

or activity, electric 97 

practical unit of electric 97 

relations of, to current, resistance and e.m.f 98 

Powers 8 

addition and subtraction of 13 

division of 10 

multiplication of 10 

of ten 12 

Practical and absolute units 116 

E.M. units directly derived 118 

units, basis of 105 

Proof of numerical value of potential 94 

Proof of the law of the angle of lag and lead 237-239 

Quantity, absolute C.G.S. electrostatic unit of 112-114 

and potential difference 203, 204 

electric 6 

magnetic, dimensions of 120 

of field 156 

potential and capacity, relation of 202, 203 

transfer of 203, 204 

Radiant fields 159 

force 33 

Rate of change 228 

of change and angle of lag 242, 243 

of change by calculus 274 

units 19 

Reactance of inductance 227-229 

capacity, ohmic equivalent of 233, 234 

of inductance, ohmic equivalent of 229 

ohmic equivalent of 227-230 

Reactances and power factor 241 

lag or lead due to 235 

Reciprocals 3 

Reduction factor 119 

Reluctance 182-185 

of air 188 

Resistance 53-76 

and e.m.f. determination of C.G.S. units of iio-iii 

and inductance, impedance of 231 



INDEX 303 

FAGB 

Resistance and temperature of conductors 73 

combined 66 

current, and e.m.f . relations of power to 98 

electrostatic unit of 115 

inductance and capacity, impedance of 236 

of a battery 82 

of a battery for maximum current by calculus 275 

of network 262, 263 

of parallel conductors 63-68 

specific 68 

Resistivity 68 

RI drop 47» 95 

Roots 8 

Roots, extraction of, by logarithms 11 

symbols of 9 

Safety fuses, calculating 130 

Saturation, chemical 143 

Sine curve 221-222 

functions 222, 223 

functions, average value of 224 

functions, average value of, by calculus 272 

functions, effective value of 225 

functions, effective value of, by calculus 273 

Space 19 

Specific inductive capacity 201, 202 

resistance 68 

Subscripts 3 

Subtraction and addition of powers 13 

algebraic 5 

Susceptibility 178-180 

table of 180 

Symbols 3 

algebraic 4 

of roots 9 

System, centimeter-gram-second ". 18 

of units, British 25 

Systems of C.G.S. electric units, two 105 

Tangent compass 108 

galvanometer formula deduced 109-110 

Temperature, absolute 137 

and resistance of conductors 73 

Theory of dimensions 35 



304 INDEX 

PAGE 

Thermo-electric chemistry 149 

couple, e.m.f. of, 131; neutral temperature of, 131; tables, 

133, 135; temperature and e.m.f. of 132 

Thomson effect 138 

Total field 188 

Traction, magnetic 192-194 

Transformers, copper loss in 216 

efi&ciency of 217 

ratio of transformation in 218 

Turns, ampere 186, 187 

ampere, for a given field 191 

Turns, C.G.S. unit 186, 187 

Unit, absolute C.G.S. electrostatic, of quantity 11 2-1 14 

field 155 

Kapp's 158 

of current, absolute C.G.S. electro-magnetic unit of 106 

of magnetic quantity 169, 170 

of moment 167 

practical, of electric energy 97 

practical, of electric power 97 

Units, absolute and practical 116 

basis of practical 105 

British system of 25 

C.G.S., of e.m.f. and resistance, how determined iio-iii 

electric, bases and relations of 105-125 

E.M. and E.S. systems of 115-116 

equivalence of 7> 28 

multiplication and division of different kinds of 17 

of E.M. system directly derived, practical 118 

rate 19 

relations of different 30-32 

simple and compound 16 

Valency 143 

Values, numerical 7 

Velocity 20 

Volt 6, 116 

Volt-coulomb 28 

Watt or volt-ampere 6, 29 

equivalents of the 100 

Watt-second 28 

Weight 22 

Weight and gravitation 18-19 

Wheatstone bridge law ; 289, 290 



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WATSON, A. E. Storage Batteries, their Theory, Construction and Use. 

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12=page Catalog of Books on Electricity, classified by 
subjects, will be furnished gratis, postage prepaid, 
on application. 



THE NEW FOSTER 

Fifth Edition, Completely Revised and Enlarged, with Four- 
fifths of Old Matter replaced by New, Up-to-date Material. 
Pocket size, flexible leather, elaborately illustrated, with an ex- 
tensive index, 1 636 pp., Thumb Index, etc. Price, $5.00 

Electrical Engineer's Pocketbook 

The Most Complete Book of Its Kind Ever Published, 

Treating of the Latest and Best Practice 

in Electrical Engineering 

By Horatio A. Foster 

Member Am. Inst. E. E., Member Am, Soc. M. E. 

With the Collaboration of Eminent Specialists 

CONTENTS 

Symbols, Units, Instruments 
Measurements 
Magnetic Properties of Iron 
Electro Magnets 
Properties of Conductors 
Relations and Dimensions of Con- 
ductors 
Underground Conduit Construction 
Standard Symbols 
Cable Testing 
Dynamos and Motors 
Tests of Dynamos and Motors 
The Static Transformer 
Standardization Rules _ 
Illuminating Engineering 
Electric I,ighting (Arc) 
Electric lyighting (Incandescent) 
Electric Street Railways 
Electrolysis 
Transmission of Power 
Storage Batteries 
Switchboards 
1,'ghtning Arresters 
Electricity Meters 
Wireless Telegraphy 
Telegraphy 
Telephony 

Electricity in the U. S. Army 
Electricity in the U. S. Navy 
Resonance 

I^ightning Conducton 
Mechanical Section 
Index 




Electric Automobiles 
Electro-chemistry and 
Electro-metallurgy 



X-Rays 

Electric Heating, Cooking 
and Welding 



D. VAN NOSTRAND COMPANY 

23 MURRAY and 27 WARREN STREETS NEW YORK 



JUN 3 1909 



!■ 



